YES Termination proof of AProVE_24_log.ari

Termination of the given RelTRS could be proven:

0 RelTRS
↳1 RelTRStoRelADPProof (⇔, 0 ms)
↳2 RelADPP
↳3 RelADPDepGraphProof (⇔, 62 ms)
↳4 AND
    ↳5 RelADPP
        ↳6 RelADPCleverAfsProof (⇒, 87 ms)
        ↳7 QDP
        ↳8 MRRProof (⇔, 2 ms)
        ↳9 QDP
        ↳10 MRRProof (⇔, 2 ms)
        ↳11 QDP
        ↳12 PisEmptyProof (⇔, 0 ms)
        ↳13 YES
    ↳14 RelADPP
        ↳15 RelADPCleverAfsProof (⇒, 59 ms)
        ↳16 QDP
        ↳17 MRRProof (⇔, 23 ms)
        ↳18 QDP
        ↳19 QDPOrderProof (⇔, 3 ms)
        ↳20 QDP
        ↳21 PisEmptyProof (⇔, 0 ms)
        ↳22 YES
    ↳23 RelADPP
        ↳24 RelADPReductionPairProof (⇔, 228 ms)
        ↳25 RelADPP
        ↳26 RelADPReductionPairProof (⇔, 44 ms)
        ↳27 RelADPP
        ↳28 DAbsisEmptyProof (⇔, 0 ms)
        ↳29 YES
    ↳30 RelADPP
        ↳31 RelADPReductionPairProof (⇔, 164 ms)
        ↳32 RelADPP
        ↳33 DAbsisEmptyProof (⇔, 0 ms)
        ↳34 YES
    ↳35 RelADPP
        ↳36 RelADPReductionPairProof (⇔, 201 ms)
        ↳37 RelADPP
        ↳38 DAbsisEmptyProof (⇔, 0 ms)
        ↳39 YES
    ↳40 RelADPP
        ↳41 RelADPReductionPairProof (⇔, 164 ms)
        ↳42 RelADPP
        ↳43 DAbsisEmptyProof (⇔, 0 ms)
        ↳44 YES
    ↳45 RelADPP
        ↳46 RelADPReductionPairProof (⇔, 269 ms)
        ↳47 RelADPP
        ↳48 DAbsisEmptyProof (⇔, 0 ms)
        ↳49 YES
    ↳50 RelADPP
        ↳51 RelADPReductionPairProof (⇔, 162 ms)
        ↳52 RelADPP
        ↳53 DAbsisEmptyProof (⇔, 0 ms)
        ↳54 YES
    ↳55 RelADPP
        ↳56 RelADPReductionPairProof (⇔, 150 ms)
        ↳57 RelADPP
        ↳58 DAbsisEmptyProof (⇔, 0 ms)
        ↳59 YES
    ↳60 RelADPP
        ↳61 RelADPReductionPairProof (⇔, 273 ms)
        ↳62 RelADPP
        ↳63 RelADPDepGraphProof (⇔, 0 ms)
        ↳64 TRUE
    ↳65 RelADPP
        ↳66 RelADPReductionPairProof (⇔, 142 ms)
        ↳67 RelADPP
        ↳68 DAbsisEmptyProof (⇔, 0 ms)
        ↳69 YES
    ↳70 RelADPP
        ↳71 RelADPReductionPairProof (⇔, 188 ms)
        ↳72 RelADPP
        ↳73 DAbsisEmptyProof (⇔, 0 ms)
        ↳74 YES
    ↳75 RelADPP
        ↳76 RelADPReductionPairProof (⇔, 159 ms)
        ↳77 RelADPP
        ↳78 DAbsisEmptyProof (⇔, 0 ms)
        ↳79 YES
    ↳80 RelADPP
        ↳81 RelADPReductionPairProof (⇔, 249 ms)
        ↳82 RelADPP
        ↳83 DAbsisEmptyProof (⇔, 0 ms)
        ↳84 YES
    ↳85 RelADPP
        ↳86 RelADPReductionPairProof (⇔, 243 ms)
        ↳87 RelADPP
        ↳88 DAbsisEmptyProof (⇔, 0 ms)
        ↳89 YES
    ↳90 RelADPP
        ↳91 RelADPReductionPairProof (⇔, 231 ms)
        ↳92 RelADPP
        ↳93 RelADPCleverAfsProof (⇒, 0 ms)
        ↳94 QDP
        ↳95 MRRProof (⇔, 0 ms)
        ↳96 QDP
        ↳97 QDPOrderProof (⇔, 0 ms)
        ↳98 QDP
        ↳99 PisEmptyProof (⇔, 0 ms)
        ↳100 YES
    ↳101 RelADPP
        ↳102 RelADPReductionPairProof (⇔, 202 ms)
        ↳103 RelADPP
        ↳104 DAbsisEmptyProof (⇔, 0 ms)
        ↳105 YES
    ↳106 RelADPP
        ↳107 RelADPReductionPairProof (⇔, 192 ms)
        ↳108 RelADPP
        ↳109 RelADPReductionPairProof (⇔, 23 ms)
        ↳110 RelADPP
        ↳111 DAbsisEmptyProof (⇔, 0 ms)
        ↳112 YES
    ↳113 RelADPP
        ↳114 RelADPReductionPairProof (⇔, 221 ms)
        ↳115 RelADPP
        ↳116 DAbsisEmptyProof (⇔, 0 ms)
        ↳117 YES
    ↳118 RelADPP
        ↳119 RelADPReductionPairProof (⇔, 301 ms)
        ↳120 RelADPP
        ↳121 DAbsisEmptyProof (⇔, 0 ms)
        ↳122 YES
    ↳123 RelADPP
        ↳124 RelADPReductionPairProof (⇔, 183 ms)
        ↳125 RelADPP
        ↳126 RelADPDepGraphProof (⇔, 0 ms)
        ↳127 TRUE
    ↳128 RelADPP
        ↳129 RelADPReductionPairProof (⇔, 152 ms)
        ↳130 RelADPP
        ↳131 DAbsisEmptyProof (⇔, 0 ms)
        ↳132 YES
    ↳133 RelADPP
        ↳134 RelADPReductionPairProof (⇔, 222 ms)
        ↳135 RelADPP
        ↳136 DAbsisEmptyProof (⇔, 0 ms)
        ↳137 YES
    ↳138 RelADPP
        ↳139 RelADPReductionPairProof (⇔, 310 ms)
        ↳140 RelADPP
        ↳141 DAbsisEmptyProof (⇔, 0 ms)
        ↳142 YES
    ↳143 RelADPP
        ↳144 RelADPReductionPairProof (⇔, 224 ms)
        ↳145 RelADPP
        ↳146 DAbsisEmptyProof (⇔, 0 ms)
        ↳147 YES
    ↳148 RelADPP
        ↳149 RelADPReductionPairProof (⇔, 199 ms)
        ↳150 RelADPP
        ↳151 RelADPReductionPairProof (⇔, 35 ms)
        ↳152 RelADPP
        ↳153 DAbsisEmptyProof (⇔, 0 ms)
        ↳154 YES
    ↳155 RelADPP
        ↳156 RelADPReductionPairProof (⇔, 268 ms)
        ↳157 RelADPP
        ↳158 DAbsisEmptyProof (⇔, 0 ms)
        ↳159 YES

(0) Obligation:

Relative term rewrite system:
The relative TRS consists of the following R rules:

minus(x, 0) → x
minus(x, s(y)) → pred(minus(x, y))
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))

The relative TRS consists of the following S rules:

s(pred(x)) → pred(s(x))
log(s(0)) → 0
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
pred(s(x)) → x

(1) RelTRStoRelADPProof (EQUIVALENT transformation)

We upgrade the RelTRS problem to an equivalent Relative ADP Problem [IJCAR24].

(2) Obligation:

Relative ADP Problem with
absolute ADPs:

minus(x, 0) → x
minus(x, s(y)) → PRED(minus(x, y))
minus(x, s(y)) → pred(MINUS(x, y))
quot(0, s(y)) → 0
quot(s(x), s(y)) → S(quot(minus(x, y), s(y)))
quot(s(x), s(y)) → s(QUOT(minus(x, y), s(y)))
quot(s(x), s(y)) → s(quot(MINUS(x, y), s(y)))
quot(s(x), s(y)) → s(quot(minus(x, y), S(y)))

and relative ADPs:

s(pred(x)) → PRED(S(x))
log(s(0)) → 0
pred(s(x)) → S(PRED(x))
log(s(s(x))) → S(LOG(s(quot(x, s(s(0))))))
log(s(s(x))) → S(log(S(quot(x, s(s(0))))))
log(s(s(x))) → S(log(s(QUOT(x, s(s(0))))))
log(s(s(x))) → S(log(s(quot(x, S(s(0))))))
log(s(s(x))) → S(log(s(quot(x, s(S(0))))))
log(s(s(x))) → s(LOG(S(quot(x, s(s(0))))))
log(s(s(x))) → s(LOG(s(QUOT(x, s(s(0))))))
log(s(s(x))) → s(LOG(s(quot(x, S(s(0))))))
log(s(s(x))) → s(LOG(s(quot(x, s(S(0))))))
log(s(s(x))) → s(log(S(QUOT(x, s(s(0))))))
log(s(s(x))) → s(log(S(quot(x, S(s(0))))))
log(s(s(x))) → s(log(S(quot(x, s(S(0))))))
log(s(s(x))) → s(log(s(QUOT(x, S(s(0))))))
log(s(s(x))) → s(log(s(QUOT(x, s(S(0))))))
log(s(s(x))) → s(log(s(quot(x, S(S(0))))))
pred(s(x)) → x

(3) RelADPDepGraphProof (EQUIVALENT transformation)

We use the relative dependency graph processor [IJCAR24].
The approximation of the Relative Dependency Graph contains:
2 SCCs with nodes from P_abs,
25 Lassos,
Result: This relative DT problem is equivalent to 27 subproblems.

(4) Complex Obligation (AND)

(5) Obligation:

Relative ADP Problem with
absolute ADPs:

minus(x, s(y)) → pred(MINUS(x, y))

and relative ADPs:

s(pred(x)) → pred(s(x))
log(s(0)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
quot(0, s(y)) → 0
pred(s(x)) → x

(6) RelADPCleverAfsProof (SOUND transformation)

We use the first derelatifying processor [IJCAR24].
There are no annotations in relative ADPs, so the relative ADP problem can be transformed into a non-relative DP problem.

Furthermore, We use an argument filter [LPAR04].
Filtering:s_1 = MINUS_2 = 0 log_1 = pred_1 = 0 = minus_2 = 1 quot_2 = 1
Found this filtering by looking at the following order that orders at least one DP strictly:Combined order from the following AFS and order.
MINUS(x1, x2)  =  MINUS(x2)
s(x1)  =  s(x1)
quot(x1, x2)  =  quot(x1)
minus(x1, x2)  =  minus(x1)
pred(x1)  =  x1
0  =  0

Recursive path order with status [RPO].
Quasi-Precedence:

MINUS1 > minus1
quot1 > s1 > minus1
0 > minus1

Status:
MINUS1: multiset
s1: [1]
quot1: [1]
minus1: [1]
0: multiset

(7) Obligation:

Q DP problem:
The TRS P consists of the following rules:

MINUS(s0(y)) → MINUS(y)

The TRS R consists of the following rules:

s0(pred0(x)) → pred0(s0(x))
log0(s0(00)) → 00
quot(s0(x)) → s0(quot(minus(x)))
minus(x) → pred0(minus(x))
pred0(s0(x)) → s0(pred0(x))
log0(s0(s0(x))) → s0(log0(s0(quot(x))))
minus(x) → x
quot(00) → 00
pred0(s0(x)) → x

Q is empty.
We have to consider all (P,Q,R)-chains.

(8) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented rules of the TRS R:

log0(s0(00)) → 00

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 2   
POL(MINUS(x1)) = x1   
POL(log0(x1)) = x1   
POL(minus(x1)) = x1   
POL(pred0(x1)) = x1   
POL(quot(x1)) = x1   
POL(s0(x1)) = 2·x1   

(9) Obligation:

Q DP problem:
The TRS P consists of the following rules:

MINUS(s0(y)) → MINUS(y)

The TRS R consists of the following rules:

s0(pred0(x)) → pred0(s0(x))
quot(s0(x)) → s0(quot(minus(x)))
minus(x) → pred0(minus(x))
pred0(s0(x)) → s0(pred0(x))
log0(s0(s0(x))) → s0(log0(s0(quot(x))))
minus(x) → x
quot(00) → 00
pred0(s0(x)) → x

Q is empty.
We have to consider all (P,Q,R)-chains.

(10) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

MINUS(s0(y)) → MINUS(y)

Strictly oriented rules of the TRS R:

pred0(s0(x)) → x

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 2   
POL(MINUS(x1)) = x1   
POL(log0(x1)) = x1   
POL(minus(x1)) = x1   
POL(pred0(x1)) = x1   
POL(quot(x1)) = x1   
POL(s0(x1)) = 1 + 2·x1   

(11) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

s0(pred0(x)) → pred0(s0(x))
quot(s0(x)) → s0(quot(minus(x)))
minus(x) → pred0(minus(x))
pred0(s0(x)) → s0(pred0(x))
log0(s0(s0(x))) → s0(log0(s0(quot(x))))
minus(x) → x
quot(00) → 00

Q is empty.
We have to consider all (P,Q,R)-chains.

(12) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(13) YES

(14) Obligation:

Relative ADP Problem with
absolute ADPs:

quot(s(x), s(y)) → s(QUOT(minus(x, y), s(y)))

and relative ADPs:

s(pred(x)) → pred(s(x))
log(s(0)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
quot(0, s(y)) → 0
pred(s(x)) → x

(15) RelADPCleverAfsProof (SOUND transformation)

We use the first derelatifying processor [IJCAR24].
There are no annotations in relative ADPs, so the relative ADP problem can be transformed into a non-relative DP problem.

Furthermore, We use an argument filter [LPAR04].
Filtering:s_1 = QUOT_2 = log_1 = pred_1 = 0 = minus_2 = 1 quot_2 =
Found this filtering by looking at the following order that orders at least one DP strictly:Combined order from the following AFS and order.
QUOT(x1, x2)  =  QUOT(x1, x2)
s(x1)  =  s(x1)
minus(x1, x2)  =  minus(x1)
pred(x1)  =  x1
0  =  0
quot(x1, x2)  =  quot(x1, x2)

Recursive path order with status [RPO].
Quasi-Precedence:

[0, quot2] > [QUOT2, s1] > minus1

Status:
QUOT2: multiset
s1: multiset
minus1: multiset
0: multiset
quot2: [2,1]

(16) Obligation:

Q DP problem:
The TRS P consists of the following rules:

QUOT0(s0(x), s0(y)) → QUOT0(minus(x), s0(y))

The TRS R consists of the following rules:

s0(pred0(x)) → pred0(s0(x))
log0(s0(00)) → 00
quot0(s0(x), s0(y)) → s0(quot0(minus(x), s0(y)))
minus(x) → pred0(minus(x))
pred0(s0(x)) → s0(pred0(x))
log0(s0(s0(x))) → s0(log0(s0(quot0(x, s0(s0(00))))))
minus(x) → x
quot0(00, s0(y)) → 00
pred0(s0(x)) → x

Q is empty.
We have to consider all (P,Q,R)-chains.

(17) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented rules of the TRS R:

log0(s0(00)) → 00

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0   
POL(QUOT0(x1, x2)) = 2·x1 + x2   
POL(log0(x1)) = 2 + 2·x1   
POL(minus(x1)) = x1   
POL(pred0(x1)) = x1   
POL(quot0(x1, x2)) = x1 + x2   
POL(s0(x1)) = x1   

(18) Obligation:

Q DP problem:
The TRS P consists of the following rules:

QUOT0(s0(x), s0(y)) → QUOT0(minus(x), s0(y))

The TRS R consists of the following rules:

s0(pred0(x)) → pred0(s0(x))
quot0(s0(x), s0(y)) → s0(quot0(minus(x), s0(y)))
minus(x) → pred0(minus(x))
pred0(s0(x)) → s0(pred0(x))
log0(s0(s0(x))) → s0(log0(s0(quot0(x, s0(s0(00))))))
minus(x) → x
quot0(00, s0(y)) → 00
pred0(s0(x)) → x

Q is empty.
We have to consider all (P,Q,R)-chains.

(19) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


QUOT0(s0(x), s0(y)) → QUOT0(minus(x), s0(y))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
QUOT0(x1, x2)  =  QUOT0(x1, x2)
s0(x1)  =  s0(x1)
minus(x1)  =  x1
pred0(x1)  =  x1
quot0(x1, x2)  =  x1
log0(x1)  =  x1
00  =  00

Recursive path order with status [RPO].
Quasi-Precedence:
QUOT02 > [s01, 00]

Status:
QUOT02: multiset
s01: multiset
00: multiset


The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

s0(pred0(x)) → pred0(s0(x))
quot0(s0(x), s0(y)) → s0(quot0(minus(x), s0(y)))
minus(x) → pred0(minus(x))
pred0(s0(x)) → s0(pred0(x))
log0(s0(s0(x))) → s0(log0(s0(quot0(x, s0(s0(00))))))
minus(x) → x
quot0(00, s0(y)) → 00
pred0(s0(x)) → x

(20) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

s0(pred0(x)) → pred0(s0(x))
quot0(s0(x), s0(y)) → s0(quot0(minus(x), s0(y)))
minus(x) → pred0(minus(x))
pred0(s0(x)) → s0(pred0(x))
log0(s0(s0(x))) → s0(log0(s0(quot0(x, s0(s0(00))))))
minus(x) → x
quot0(00, s0(y)) → 00
pred0(s0(x)) → x

Q is empty.
We have to consider all (P,Q,R)-chains.

(21) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(22) YES

(23) Obligation:

Relative ADP Problem with
absolute ADPs:

quot(s(x), s(y)) → s(QUOT(minus(x, y), s(y)))

and relative ADPs:

log(s(s(x))) → s(LOG(s(quot(x, s(S(0))))))
log(s(0)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(s(s(x))) → s(LOG(s(QUOT(x, s(s(0))))))
log(s(s(x))) → s(LOG(S(quot(x, s(s(0))))))
quot(0, s(y)) → 0
pred(s(x)) → x
log(s(s(x))) → s(LOG(s(quot(x, S(s(0))))))
s(pred(x)) → pred(s(x))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
log(s(s(x))) → s(log(S(QUOT(x, s(s(0))))))
minus(x, 0) → x
log(s(s(x))) → S(LOG(s(quot(x, s(s(0))))))

(24) RelADPReductionPairProof (EQUIVALENT transformation)

We use the reduction pair processor [IJCAR24].
The following rules can be oriented strictly (l^# > ann(r))
and therefore we can remove all of its annotations in the right-hand side:
Relative ADPs:


log(s(0)) → 0
s(pred(x)) → pred(s(x))
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
log(s(s(x))) → s(log(S(QUOT(x, s(s(0))))))
minus(x, 0) → x
quot(0, s(y)) → 0
pred(s(x)) → x


The remaining rules can at least be oriented weakly:
Absolute ADPs:

quot(s(x), s(y)) → s(QUOT(minus(x, y), s(y)))

Relative ADPs:

log(s(s(x))) → s(LOG(s(quot(x, s(S(0))))))
log(s(s(x))) → s(LOG(s(quot(x, S(s(0))))))
log(s(s(x))) → s(LOG(s(QUOT(x, s(s(0))))))
log(s(s(x))) → s(LOG(S(quot(x, s(s(0))))))
log(s(s(x))) → S(LOG(s(quot(x, s(s(0))))))


Ordered with Polynomial interpretation [POLO]:

POL(0) = 0   
POL(LOG(x1)) = 2 + x1   
POL(MINUS(x1, x2)) = 2 + 2·x1 + 2·x2   
POL(PRED(x1)) = 2   
POL(QUOT(x1, x2)) = 0   
POL(S(x1)) = 0   
POL(log(x1)) = 2·x1   
POL(minus(x1, x2)) = x1   
POL(pred(x1)) = x1   
POL(quot(x1, x2)) = x2   
POL(s(x1)) = x1   

(25) Obligation:

Relative ADP Problem with
absolute ADPs:

quot(s(x), s(y)) → s(QUOT(minus(x, y), s(y)))

and relative ADPs:

log(s(s(x))) → s(LOG(s(quot(x, s(S(0))))))
log(s(0)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(s(s(x))) → s(LOG(s(QUOT(x, s(s(0))))))
log(s(s(x))) → s(LOG(S(quot(x, s(s(0))))))
quot(0, s(y)) → 0
pred(s(x)) → x
log(s(s(x))) → s(LOG(s(quot(x, S(s(0))))))
s(pred(x)) → pred(s(x))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
log(s(s(x))) → S(LOG(s(quot(x, s(s(0))))))

(26) RelADPReductionPairProof (EQUIVALENT transformation)

We use the reduction pair processor [IJCAR24].
The following rules can be oriented strictly (l^# > ann(r))
and therefore we can remove all of its annotations in the right-hand side:
Absolute ADPs:


quot(s(x), s(y)) → s(QUOT(minus(x, y), s(y)))

Relative ADPs:

log(s(s(x))) → s(LOG(s(quot(x, s(S(0))))))
log(s(0)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(s(s(x))) → s(LOG(s(QUOT(x, s(s(0))))))
log(s(s(x))) → s(LOG(S(quot(x, s(s(0))))))
quot(0, s(y)) → 0
pred(s(x)) → x
log(s(s(x))) → s(LOG(s(quot(x, S(s(0))))))
s(pred(x)) → pred(s(x))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
log(s(s(x))) → S(LOG(s(quot(x, s(s(0))))))


No rules with annotations remain.
Ordered with Polynomial interpretation [POLO]:

POL(0) = 0   
POL(LOG(x1)) = 2 + 2·x1   
POL(MINUS(x1, x2)) = 2 + 2·x1 + 2·x1·x2 + 2·x2   
POL(PRED(x1)) = 2   
POL(QUOT(x1, x2)) = 3·x1 + x2   
POL(S(x1)) = 3   
POL(log(x1)) = 2 + 3·x1   
POL(minus(x1, x2)) = x1   
POL(pred(x1)) = x1   
POL(quot(x1, x2)) = x1   
POL(s(x1)) = 3 + 3·x1   

(27) Obligation:

Relative ADP Problem with
No absolute ADPs, and relative ADPs:

log(s(0)) → 0
s(pred(x)) → pred(s(x))
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
quot(0, s(y)) → 0
pred(s(x)) → x

(28) DAbsisEmptyProof (EQUIVALENT transformation)

The relative ADP Problem has an empty P_abs. Hence, no infinite chain exists.

(29) YES

(30) Obligation:

Relative ADP Problem with
absolute ADPs:

quot(s(x), s(y)) → S(quot(minus(x, y), s(y)))

and relative ADPs:

log(s(s(x))) → s(LOG(s(quot(x, s(S(0))))))
log(s(0)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(s(s(x))) → s(LOG(s(QUOT(x, s(s(0))))))
log(s(s(x))) → s(LOG(S(quot(x, s(s(0))))))
quot(0, s(y)) → 0
pred(s(x)) → x
log(s(s(x))) → s(LOG(s(quot(x, S(s(0))))))
s(pred(x)) → pred(s(x))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
log(s(s(x))) → s(log(S(QUOT(x, s(s(0))))))
minus(x, 0) → x
log(s(s(x))) → S(LOG(s(quot(x, s(s(0))))))

(31) RelADPReductionPairProof (EQUIVALENT transformation)

We use the reduction pair processor [IJCAR24].
The following rules can be oriented strictly (l^# > ann(r))
and therefore we can remove all of its annotations in the right-hand side:
Absolute ADPs:


quot(s(x), s(y)) → S(quot(minus(x, y), s(y)))

Relative ADPs:

log(s(s(x))) → s(LOG(s(quot(x, s(S(0))))))
log(s(0)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(s(s(x))) → s(LOG(s(QUOT(x, s(s(0))))))
log(s(s(x))) → s(LOG(S(quot(x, s(s(0))))))
quot(0, s(y)) → 0
pred(s(x)) → x
s(pred(x)) → pred(s(x))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
log(s(s(x))) → s(log(S(QUOT(x, s(s(0))))))
minus(x, 0) → x
log(s(s(x))) → S(LOG(s(quot(x, s(s(0))))))


The remaining rules can at least be oriented weakly:

Ordered with Polynomial interpretation [POLO]:

POL(0) = 2   
POL(LOG(x1)) = 3 + 3·x1   
POL(MINUS(x1, x2)) = 2 + 2·x1   
POL(PRED(x1)) = 2   
POL(QUOT(x1, x2)) = 1 + 3·x1   
POL(S(x1)) = 3·x1   
POL(log(x1)) = 2·x1   
POL(minus(x1, x2)) = x1   
POL(pred(x1)) = x1   
POL(quot(x1, x2)) = x1   
POL(s(x1)) = 3 + 3·x1   

(32) Obligation:

Relative ADP Problem with
No absolute ADPs, and relative ADPs:

log(s(s(x))) → s(LOG(s(quot(x, S(s(0))))))
log(s(0)) → 0
s(pred(x)) → pred(s(x))
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
quot(0, s(y)) → 0
pred(s(x)) → x

(33) DAbsisEmptyProof (EQUIVALENT transformation)

The relative ADP Problem has an empty P_abs. Hence, no infinite chain exists.

(34) YES

(35) Obligation:

Relative ADP Problem with
absolute ADPs:

quot(0, s(y)) → 0

and relative ADPs:

log(s(s(x))) → s(LOG(s(quot(x, s(S(0))))))
log(s(0)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(s(s(x))) → s(LOG(s(QUOT(x, s(s(0))))))
log(s(s(x))) → s(LOG(S(quot(x, s(s(0))))))
log(s(s(x))) → s(log(s(QUOT(x, S(s(0))))))
pred(s(x)) → x
log(s(s(x))) → s(LOG(s(quot(x, S(s(0))))))
s(pred(x)) → pred(s(x))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
log(s(s(x))) → S(LOG(s(quot(x, s(s(0))))))

(36) RelADPReductionPairProof (EQUIVALENT transformation)

We use the reduction pair processor [IJCAR24].
The following rules can be oriented strictly (l^# > ann(r))
and therefore we can remove all of its annotations in the right-hand side:
Absolute ADPs:


quot(0, s(y)) → 0

Relative ADPs:

log(s(s(x))) → s(LOG(s(quot(x, s(S(0))))))
log(s(0)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(s(s(x))) → s(LOG(S(quot(x, s(s(0))))))
log(s(s(x))) → s(log(s(QUOT(x, S(s(0))))))
pred(s(x)) → x
log(s(s(x))) → s(LOG(s(quot(x, S(s(0))))))
s(pred(x)) → pred(s(x))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
log(s(s(x))) → S(LOG(s(quot(x, s(s(0))))))


The remaining rules can at least be oriented weakly:

Ordered with Polynomial interpretation [POLO]:

POL(0) = 0   
POL(LOG(x1)) = 3 + 3·x1   
POL(MINUS(x1, x2)) = 2·x1   
POL(PRED(x1)) = 2   
POL(QUOT(x1, x2)) = 2·x1 + 2·x2   
POL(S(x1)) = 3 + 2·x1   
POL(log(x1)) = x1   
POL(minus(x1, x2)) = x1   
POL(pred(x1)) = x1   
POL(quot(x1, x2)) = x1   
POL(s(x1)) = 3 + 2·x1   

(37) Obligation:

Relative ADP Problem with
No absolute ADPs, and relative ADPs:

log(s(0)) → 0
s(pred(x)) → pred(s(x))
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(s(s(x))) → s(LOG(s(QUOT(x, s(s(0))))))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
quot(0, s(y)) → 0
pred(s(x)) → x

(38) DAbsisEmptyProof (EQUIVALENT transformation)

The relative ADP Problem has an empty P_abs. Hence, no infinite chain exists.

(39) YES

(40) Obligation:

Relative ADP Problem with
absolute ADPs:

quot(s(x), s(y)) → s(quot(MINUS(x, y), s(y)))

and relative ADPs:

log(s(s(x))) → s(LOG(s(quot(x, s(S(0))))))
log(s(0)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(s(s(x))) → s(LOG(s(QUOT(x, s(s(0))))))
log(s(s(x))) → s(LOG(S(quot(x, s(s(0))))))
log(s(s(x))) → s(log(s(QUOT(x, S(s(0))))))
quot(0, s(y)) → 0
pred(s(x)) → x
log(s(s(x))) → s(LOG(s(quot(x, S(s(0))))))
s(pred(x)) → pred(s(x))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
log(s(s(x))) → S(LOG(s(quot(x, s(s(0))))))

(41) RelADPReductionPairProof (EQUIVALENT transformation)

We use the reduction pair processor [IJCAR24].
The following rules can be oriented strictly (l^# > ann(r))
and therefore we can remove all of its annotations in the right-hand side:
Absolute ADPs:


quot(s(x), s(y)) → s(quot(MINUS(x, y), s(y)))

Relative ADPs:

log(s(s(x))) → s(LOG(s(quot(x, s(S(0))))))
log(s(0)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(s(s(x))) → s(LOG(s(QUOT(x, s(s(0))))))
log(s(s(x))) → s(LOG(S(quot(x, s(s(0))))))
log(s(s(x))) → s(log(s(QUOT(x, S(s(0))))))
quot(0, s(y)) → 0
pred(s(x)) → x
log(s(s(x))) → s(LOG(s(quot(x, S(s(0))))))
s(pred(x)) → pred(s(x))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
log(s(s(x))) → S(LOG(s(quot(x, s(s(0))))))


No rules with annotations remain.
Ordered with Polynomial interpretation [POLO]:

POL(0) = 0   
POL(LOG(x1)) = 2 + 2·x1   
POL(MINUS(x1, x2)) = 1 + 3·x2   
POL(PRED(x1)) = 2   
POL(QUOT(x1, x2)) = 3 + x2   
POL(S(x1)) = 2 + x1   
POL(log(x1)) = 3 + 3·x1   
POL(minus(x1, x2)) = x1   
POL(pred(x1)) = x1   
POL(quot(x1, x2)) = x1   
POL(s(x1)) = 3 + 3·x1   

(42) Obligation:

Relative ADP Problem with
No absolute ADPs, and relative ADPs:

log(s(0)) → 0
s(pred(x)) → pred(s(x))
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
quot(0, s(y)) → 0
pred(s(x)) → x

(43) DAbsisEmptyProof (EQUIVALENT transformation)

The relative ADP Problem has an empty P_abs. Hence, no infinite chain exists.

(44) YES

(45) Obligation:

Relative ADP Problem with
absolute ADPs:

quot(0, s(y)) → 0

and relative ADPs:

log(s(s(x))) → s(LOG(s(quot(x, s(S(0))))))
log(s(0)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(s(s(x))) → s(LOG(s(QUOT(x, s(s(0))))))
log(s(s(x))) → s(LOG(S(quot(x, s(s(0))))))
pred(s(x)) → x
log(s(s(x))) → s(LOG(s(quot(x, S(s(0))))))
s(pred(x)) → pred(s(x))
log(s(s(x))) → S(log(s(QUOT(x, s(s(0))))))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
log(s(s(x))) → S(LOG(s(quot(x, s(s(0))))))

(46) RelADPReductionPairProof (EQUIVALENT transformation)

We use the reduction pair processor [IJCAR24].
The following rules can be oriented strictly (l^# > ann(r))
and therefore we can remove all of its annotations in the right-hand side:
Absolute ADPs:


quot(0, s(y)) → 0

Relative ADPs:

log(s(s(x))) → s(LOG(s(quot(x, s(S(0))))))
log(s(0)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(s(s(x))) → s(LOG(s(QUOT(x, s(s(0))))))
log(s(s(x))) → s(LOG(S(quot(x, s(s(0))))))
pred(s(x)) → x
log(s(s(x))) → s(LOG(s(quot(x, S(s(0))))))
s(pred(x)) → pred(s(x))
log(s(s(x))) → S(log(s(QUOT(x, s(s(0))))))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
log(s(s(x))) → S(LOG(s(quot(x, s(s(0))))))


No rules with annotations remain.
Ordered with Polynomial interpretation [POLO]:

POL(0) = 0   
POL(LOG(x1)) = 3 + 3·x1   
POL(MINUS(x1, x2)) = 2 + 2·x1   
POL(PRED(x1)) = 2   
POL(QUOT(x1, x2)) = 3   
POL(S(x1)) = 3 + 2·x1   
POL(log(x1)) = x1   
POL(minus(x1, x2)) = x1   
POL(pred(x1)) = x1   
POL(quot(x1, x2)) = x1   
POL(s(x1)) = 3 + 2·x1   

(47) Obligation:

Relative ADP Problem with
No absolute ADPs, and relative ADPs:

log(s(0)) → 0
s(pred(x)) → pred(s(x))
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
quot(0, s(y)) → 0
pred(s(x)) → x

(48) DAbsisEmptyProof (EQUIVALENT transformation)

The relative ADP Problem has an empty P_abs. Hence, no infinite chain exists.

(49) YES

(50) Obligation:

Relative ADP Problem with
absolute ADPs:

quot(s(x), s(y)) → s(quot(MINUS(x, y), s(y)))

and relative ADPs:

log(s(s(x))) → s(LOG(s(quot(x, s(S(0))))))
log(s(0)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(s(s(x))) → s(LOG(s(QUOT(x, s(s(0))))))
log(s(s(x))) → s(LOG(S(quot(x, s(s(0))))))
quot(0, s(y)) → 0
pred(s(x)) → x
log(s(s(x))) → s(LOG(s(quot(x, S(s(0))))))
s(pred(x)) → pred(s(x))
log(s(s(x))) → S(log(s(QUOT(x, s(s(0))))))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
log(s(s(x))) → S(LOG(s(quot(x, s(s(0))))))

(51) RelADPReductionPairProof (EQUIVALENT transformation)

We use the reduction pair processor [IJCAR24].
The following rules can be oriented strictly (l^# > ann(r))
and therefore we can remove all of its annotations in the right-hand side:
Absolute ADPs:


quot(s(x), s(y)) → s(quot(MINUS(x, y), s(y)))

Relative ADPs:

log(s(s(x))) → s(LOG(s(quot(x, s(S(0))))))
log(s(0)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(s(s(x))) → s(LOG(s(QUOT(x, s(s(0))))))
log(s(s(x))) → s(LOG(S(quot(x, s(s(0))))))
quot(0, s(y)) → 0
pred(s(x)) → x
log(s(s(x))) → s(LOG(s(quot(x, S(s(0))))))
s(pred(x)) → pred(s(x))
log(s(s(x))) → S(log(s(QUOT(x, s(s(0))))))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
log(s(s(x))) → S(LOG(s(quot(x, s(s(0))))))


No rules with annotations remain.
Ordered with Polynomial interpretation [POLO]:

POL(0) = 0   
POL(LOG(x1)) = 3 + 2·x1   
POL(MINUS(x1, x2)) = 0   
POL(PRED(x1)) = 2   
POL(QUOT(x1, x2)) = 2   
POL(S(x1)) = 1   
POL(log(x1)) = 3·x1   
POL(minus(x1, x2)) = x1   
POL(pred(x1)) = x1   
POL(quot(x1, x2)) = 1 + x1   
POL(s(x1)) = 3 + x1   

(52) Obligation:

Relative ADP Problem with
No absolute ADPs, and relative ADPs:

log(s(0)) → 0
s(pred(x)) → pred(s(x))
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
quot(0, s(y)) → 0
pred(s(x)) → x

(53) DAbsisEmptyProof (EQUIVALENT transformation)

The relative ADP Problem has an empty P_abs. Hence, no infinite chain exists.

(54) YES

(55) Obligation:

Relative ADP Problem with
absolute ADPs:

quot(s(x), s(y)) → S(quot(minus(x, y), s(y)))

and relative ADPs:

log(s(s(x))) → s(LOG(s(quot(x, s(S(0))))))
log(s(0)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(s(s(x))) → s(LOG(s(QUOT(x, s(s(0))))))
log(s(s(x))) → s(LOG(S(quot(x, s(s(0))))))
log(s(s(x))) → s(log(s(QUOT(x, S(s(0))))))
quot(0, s(y)) → 0
pred(s(x)) → x
log(s(s(x))) → s(LOG(s(quot(x, S(s(0))))))
s(pred(x)) → pred(s(x))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
log(s(s(x))) → S(LOG(s(quot(x, s(s(0))))))

(56) RelADPReductionPairProof (EQUIVALENT transformation)

We use the reduction pair processor [IJCAR24].
The following rules can be oriented strictly (l^# > ann(r))
and therefore we can remove all of its annotations in the right-hand side:
Absolute ADPs:


quot(s(x), s(y)) → S(quot(minus(x, y), s(y)))

Relative ADPs:

log(s(s(x))) → s(LOG(s(quot(x, s(S(0))))))
log(s(0)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(s(s(x))) → s(LOG(s(QUOT(x, s(s(0))))))
log(s(s(x))) → s(LOG(S(quot(x, s(s(0))))))
log(s(s(x))) → s(log(s(QUOT(x, S(s(0))))))
quot(0, s(y)) → 0
pred(s(x)) → x
log(s(s(x))) → s(LOG(s(quot(x, S(s(0))))))
s(pred(x)) → pred(s(x))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
log(s(s(x))) → S(LOG(s(quot(x, s(s(0))))))


No rules with annotations remain.
Ordered with Polynomial interpretation [POLO]:

POL(0) = 0   
POL(LOG(x1)) = 3 + 3·x1   
POL(MINUS(x1, x2)) = 2 + 2·x1 + 2·x2   
POL(PRED(x1)) = 2 + 2·x12   
POL(QUOT(x1, x2)) = 3·x1 + x2   
POL(S(x1)) = 2 + x1   
POL(log(x1)) = 2·x1   
POL(minus(x1, x2)) = x1   
POL(pred(x1)) = x1   
POL(quot(x1, x2)) = x1   
POL(s(x1)) = 3 + 2·x1   

(57) Obligation:

Relative ADP Problem with
No absolute ADPs, and relative ADPs:

log(s(0)) → 0
s(pred(x)) → pred(s(x))
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
quot(0, s(y)) → 0
pred(s(x)) → x

(58) DAbsisEmptyProof (EQUIVALENT transformation)

The relative ADP Problem has an empty P_abs. Hence, no infinite chain exists.

(59) YES

(60) Obligation:

Relative ADP Problem with
absolute ADPs:

quot(s(x), s(y)) → s(quot(minus(x, y), S(y)))

and relative ADPs:

log(s(s(x))) → s(LOG(s(quot(x, s(S(0))))))
log(s(0)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(s(s(x))) → s(LOG(s(QUOT(x, s(s(0))))))
log(s(s(x))) → s(LOG(S(quot(x, s(s(0))))))
quot(0, s(y)) → 0
pred(s(x)) → x
log(s(s(x))) → s(LOG(s(quot(x, S(s(0))))))
s(pred(x)) → pred(s(x))
log(s(s(x))) → S(log(s(QUOT(x, s(s(0))))))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
log(s(s(x))) → S(LOG(s(quot(x, s(s(0))))))

(61) RelADPReductionPairProof (EQUIVALENT transformation)

We use the reduction pair processor [IJCAR24].
The following rules can be oriented strictly (l^# > ann(r))
and therefore we can remove all of its annotations in the right-hand side:
Relative ADPs:


log(s(s(x))) → s(LOG(s(quot(x, s(S(0))))))
log(s(0)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(s(s(x))) → s(LOG(s(QUOT(x, s(s(0))))))
log(s(s(x))) → s(LOG(S(quot(x, s(s(0))))))
quot(0, s(y)) → 0
pred(s(x)) → x
log(s(s(x))) → s(LOG(s(quot(x, S(s(0))))))
s(pred(x)) → pred(s(x))
log(s(s(x))) → S(log(s(QUOT(x, s(s(0))))))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
log(s(s(x))) → S(LOG(s(quot(x, s(s(0))))))


The remaining rules can at least be oriented weakly:
Absolute ADPs:

quot(s(x), s(y)) → s(quot(minus(x, y), S(y)))

Relative ADPs:
none


Ordered with Polynomial interpretation [POLO]:

POL(0) = 0   
POL(LOG(x1)) = 2 + 3·x1   
POL(MINUS(x1, x2)) = 2 + 2·x1   
POL(PRED(x1)) = x12   
POL(QUOT(x1, x2)) = x2   
POL(S(x1)) = 3 + x1   
POL(log(x1)) = 1 + 2·x1   
POL(minus(x1, x2)) = x1   
POL(pred(x1)) = x1   
POL(quot(x1, x2)) = x1   
POL(s(x1)) = 3 + 3·x1   

(62) Obligation:

Relative ADP Problem with
absolute ADPs:

quot(s(x), s(y)) → s(quot(minus(x, y), S(y)))

and relative ADPs:

log(s(0)) → 0
s(pred(x)) → pred(s(x))
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
quot(0, s(y)) → 0
pred(s(x)) → x

(63) RelADPDepGraphProof (EQUIVALENT transformation)

We use the relative dependency graph processor [IJCAR24].
The approximation of the Relative Dependency Graph contains:
0 SCCs with nodes from P_abs,
0 Lassos,
Result: This relative DT problem is equivalent to 0 subproblems.

(64) TRUE

(65) Obligation:

Relative ADP Problem with
absolute ADPs:

quot(0, s(y)) → 0

and relative ADPs:

log(s(s(x))) → s(LOG(s(quot(x, s(S(0))))))
log(s(0)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(s(s(x))) → s(LOG(s(QUOT(x, s(s(0))))))
log(s(s(x))) → s(LOG(S(quot(x, s(s(0))))))
pred(s(x)) → x
log(s(s(x))) → s(LOG(s(quot(x, S(s(0))))))
s(pred(x)) → pred(s(x))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
log(s(s(x))) → S(LOG(s(quot(x, s(s(0))))))

(66) RelADPReductionPairProof (EQUIVALENT transformation)

We use the reduction pair processor [IJCAR24].
The following rules can be oriented strictly (l^# > ann(r))
and therefore we can remove all of its annotations in the right-hand side:
Absolute ADPs:


quot(0, s(y)) → 0

Relative ADPs:

log(s(s(x))) → s(LOG(s(quot(x, s(S(0))))))
log(s(s(x))) → s(LOG(s(quot(x, S(s(0))))))
log(s(0)) → 0
s(pred(x)) → pred(s(x))
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(s(s(x))) → s(LOG(s(QUOT(x, s(s(0))))))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
log(s(s(x))) → s(LOG(S(quot(x, s(s(0))))))
pred(s(x)) → x


The remaining rules can at least be oriented weakly:

Ordered with Polynomial interpretation [POLO]:

POL(0) = 0   
POL(LOG(x1)) = 2 + 2·x1   
POL(MINUS(x1, x2)) = 2 + 2·x1·x2   
POL(PRED(x1)) = 2 + x12   
POL(QUOT(x1, x2)) = 1 + x2   
POL(S(x1)) = 3 + x1   
POL(log(x1)) = 3·x1   
POL(minus(x1, x2)) = x1   
POL(pred(x1)) = x1   
POL(quot(x1, x2)) = x1   
POL(s(x1)) = 1 + 3·x1   

(67) Obligation:

Relative ADP Problem with
No absolute ADPs, and relative ADPs:

log(s(0)) → 0
s(pred(x)) → pred(s(x))
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
log(s(s(x))) → S(LOG(s(quot(x, s(s(0))))))
quot(0, s(y)) → 0
pred(s(x)) → x

(68) DAbsisEmptyProof (EQUIVALENT transformation)

The relative ADP Problem has an empty P_abs. Hence, no infinite chain exists.

(69) YES

(70) Obligation:

Relative ADP Problem with
absolute ADPs:

quot(s(x), s(y)) → s(quot(minus(x, y), S(y)))

and relative ADPs:

log(s(s(x))) → s(LOG(s(quot(x, s(S(0))))))
log(s(0)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(s(s(x))) → s(LOG(s(QUOT(x, s(s(0))))))
log(s(s(x))) → s(LOG(S(quot(x, s(s(0))))))
quot(0, s(y)) → 0
pred(s(x)) → x
log(s(s(x))) → s(LOG(s(quot(x, S(s(0))))))
s(pred(x)) → pred(s(x))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
log(s(s(x))) → S(LOG(s(quot(x, s(s(0))))))

(71) RelADPReductionPairProof (EQUIVALENT transformation)

We use the reduction pair processor [IJCAR24].
The following rules can be oriented strictly (l^# > ann(r))
and therefore we can remove all of its annotations in the right-hand side:
Absolute ADPs:


quot(s(x), s(y)) → s(quot(minus(x, y), S(y)))

Relative ADPs:

log(s(s(x))) → s(LOG(s(quot(x, s(S(0))))))
log(s(0)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(s(s(x))) → s(LOG(s(QUOT(x, s(s(0))))))
log(s(s(x))) → s(LOG(S(quot(x, s(s(0))))))
quot(0, s(y)) → 0
pred(s(x)) → x
log(s(s(x))) → s(LOG(s(quot(x, S(s(0))))))
s(pred(x)) → pred(s(x))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
log(s(s(x))) → S(LOG(s(quot(x, s(s(0))))))


No rules with annotations remain.
Ordered with Polynomial interpretation [POLO]:

POL(0) = 0   
POL(LOG(x1)) = 2 + 2·x1   
POL(MINUS(x1, x2)) = 2 + 2·x1 + 2·x2   
POL(PRED(x1)) = 2   
POL(QUOT(x1, x2)) = 2 + x2   
POL(S(x1)) = 2 + 2·x1   
POL(log(x1)) = x1   
POL(minus(x1, x2)) = x1   
POL(pred(x1)) = x1   
POL(quot(x1, x2)) = x1   
POL(s(x1)) = 3 + 3·x1   

(72) Obligation:

Relative ADP Problem with
No absolute ADPs, and relative ADPs:

log(s(0)) → 0
s(pred(x)) → pred(s(x))
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
quot(0, s(y)) → 0
pred(s(x)) → x

(73) DAbsisEmptyProof (EQUIVALENT transformation)

The relative ADP Problem has an empty P_abs. Hence, no infinite chain exists.

(74) YES

(75) Obligation:

Relative ADP Problem with
absolute ADPs:

quot(s(x), s(y)) → s(quot(MINUS(x, y), s(y)))

and relative ADPs:

log(s(s(x))) → s(LOG(s(quot(x, s(S(0))))))
log(s(0)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(s(s(x))) → s(LOG(s(QUOT(x, s(s(0))))))
log(s(s(x))) → s(LOG(S(quot(x, s(s(0))))))
quot(0, s(y)) → 0
pred(s(x)) → x
log(s(s(x))) → s(LOG(s(quot(x, S(s(0))))))
s(pred(x)) → pred(s(x))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
log(s(s(x))) → S(LOG(s(quot(x, s(s(0))))))

(76) RelADPReductionPairProof (EQUIVALENT transformation)

We use the reduction pair processor [IJCAR24].
The following rules can be oriented strictly (l^# > ann(r))
and therefore we can remove all of its annotations in the right-hand side:
Absolute ADPs:


quot(s(x), s(y)) → s(quot(MINUS(x, y), s(y)))

Relative ADPs:

log(s(s(x))) → s(LOG(s(quot(x, s(S(0))))))
log(s(0)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(s(s(x))) → s(LOG(s(QUOT(x, s(s(0))))))
log(s(s(x))) → s(LOG(S(quot(x, s(s(0))))))
quot(0, s(y)) → 0
pred(s(x)) → x
log(s(s(x))) → s(LOG(s(quot(x, S(s(0))))))
s(pred(x)) → pred(s(x))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x


The remaining rules can at least be oriented weakly:

Ordered with Polynomial interpretation [POLO]:

POL(0) = 0   
POL(LOG(x1)) = 2 + 2·x1   
POL(MINUS(x1, x2)) = 0   
POL(PRED(x1)) = 2   
POL(QUOT(x1, x2)) = 3   
POL(S(x1)) = 3 + x1   
POL(log(x1)) = 3·x1   
POL(minus(x1, x2)) = x1   
POL(pred(x1)) = x1   
POL(quot(x1, x2)) = x1   
POL(s(x1)) = 1 + 3·x1   

(77) Obligation:

Relative ADP Problem with
No absolute ADPs, and relative ADPs:

log(s(0)) → 0
s(pred(x)) → pred(s(x))
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
log(s(s(x))) → S(LOG(s(quot(x, s(s(0))))))
quot(0, s(y)) → 0
pred(s(x)) → x

(78) DAbsisEmptyProof (EQUIVALENT transformation)

The relative ADP Problem has an empty P_abs. Hence, no infinite chain exists.

(79) YES

(80) Obligation:

Relative ADP Problem with
absolute ADPs:

quot(s(x), s(y)) → s(quot(minus(x, y), S(y)))

and relative ADPs:

log(s(s(x))) → s(LOG(s(quot(x, s(S(0))))))
log(s(0)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(s(s(x))) → s(LOG(s(QUOT(x, s(s(0))))))
log(s(s(x))) → s(LOG(S(quot(x, s(s(0))))))
quot(0, s(y)) → 0
pred(s(x)) → x
log(s(s(x))) → s(LOG(s(quot(x, S(s(0))))))
s(pred(x)) → pred(s(x))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
log(s(s(x))) → s(log(S(QUOT(x, s(s(0))))))
minus(x, 0) → x
log(s(s(x))) → S(LOG(s(quot(x, s(s(0))))))

(81) RelADPReductionPairProof (EQUIVALENT transformation)

We use the reduction pair processor [IJCAR24].
The following rules can be oriented strictly (l^# > ann(r))
and therefore we can remove all of its annotations in the right-hand side:
Absolute ADPs:


quot(s(x), s(y)) → s(quot(minus(x, y), S(y)))

Relative ADPs:

log(s(s(x))) → s(LOG(s(quot(x, s(S(0))))))
log(s(0)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(s(s(x))) → s(LOG(s(QUOT(x, s(s(0))))))
log(s(s(x))) → s(LOG(S(quot(x, s(s(0))))))
quot(0, s(y)) → 0
pred(s(x)) → x
log(s(s(x))) → s(LOG(s(quot(x, S(s(0))))))
s(pred(x)) → pred(s(x))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
log(s(s(x))) → s(log(S(QUOT(x, s(s(0))))))
minus(x, 0) → x


The remaining rules can at least be oriented weakly:

Ordered with Polynomial interpretation [POLO]:

POL(0) = 0   
POL(LOG(x1)) = 2 + 3·x1   
POL(MINUS(x1, x2)) = 2 + 2·x1   
POL(PRED(x1)) = x12   
POL(QUOT(x1, x2)) = 3 + 2·x1 + x2   
POL(S(x1)) = 3 + 2·x1   
POL(log(x1)) = 3 + 3·x1   
POL(minus(x1, x2)) = x1   
POL(pred(x1)) = x1   
POL(quot(x1, x2)) = x1   
POL(s(x1)) = 3 + 3·x1   

(82) Obligation:

Relative ADP Problem with
No absolute ADPs, and relative ADPs:

log(s(0)) → 0
s(pred(x)) → pred(s(x))
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
log(s(s(x))) → S(LOG(s(quot(x, s(s(0))))))
quot(0, s(y)) → 0
pred(s(x)) → x

(83) DAbsisEmptyProof (EQUIVALENT transformation)

The relative ADP Problem has an empty P_abs. Hence, no infinite chain exists.

(84) YES

(85) Obligation:

Relative ADP Problem with
absolute ADPs:

quot(s(x), s(y)) → S(quot(minus(x, y), s(y)))

and relative ADPs:

log(s(s(x))) → s(LOG(s(quot(x, s(S(0))))))
log(s(0)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(s(s(x))) → s(LOG(s(QUOT(x, s(s(0))))))
log(s(s(x))) → s(LOG(S(quot(x, s(s(0))))))
quot(0, s(y)) → 0
pred(s(x)) → x
log(s(s(x))) → s(LOG(s(quot(x, S(s(0))))))
s(pred(x)) → pred(s(x))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
log(s(s(x))) → S(LOG(s(quot(x, s(s(0))))))
log(s(s(x))) → s(log(s(QUOT(x, s(S(0))))))

(86) RelADPReductionPairProof (EQUIVALENT transformation)

We use the reduction pair processor [IJCAR24].
The following rules can be oriented strictly (l^# > ann(r))
and therefore we can remove all of its annotations in the right-hand side:
Absolute ADPs:


quot(s(x), s(y)) → S(quot(minus(x, y), s(y)))

Relative ADPs:

log(s(s(x))) → s(LOG(s(quot(x, s(S(0))))))
log(s(0)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(s(s(x))) → s(LOG(S(quot(x, s(s(0))))))
quot(0, s(y)) → 0
pred(s(x)) → x
log(s(s(x))) → s(LOG(s(quot(x, S(s(0))))))
s(pred(x)) → pred(s(x))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
log(s(s(x))) → S(LOG(s(quot(x, s(s(0))))))
log(s(s(x))) → s(log(s(QUOT(x, s(S(0))))))


The remaining rules can at least be oriented weakly:

Ordered with Polynomial interpretation [POLO]:

POL(0) = 0   
POL(LOG(x1)) = 1 + 3·x1   
POL(MINUS(x1, x2)) = 2 + 2·x1 + 2·x1·x2   
POL(PRED(x1)) = 0   
POL(QUOT(x1, x2)) = 3   
POL(S(x1)) = 1   
POL(log(x1)) = 3·x1   
POL(minus(x1, x2)) = x1   
POL(pred(x1)) = x1   
POL(quot(x1, x2)) = 2 + x1   
POL(s(x1)) = 3 + x1   

(87) Obligation:

Relative ADP Problem with
No absolute ADPs, and relative ADPs:

log(s(0)) → 0
s(pred(x)) → pred(s(x))
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(s(s(x))) → s(LOG(s(QUOT(x, s(s(0))))))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
quot(0, s(y)) → 0
pred(s(x)) → x

(88) DAbsisEmptyProof (EQUIVALENT transformation)

The relative ADP Problem has an empty P_abs. Hence, no infinite chain exists.

(89) YES

(90) Obligation:

Relative ADP Problem with
absolute ADPs:

quot(s(x), s(y)) → s(QUOT(minus(x, y), s(y)))

and relative ADPs:

log(s(s(x))) → s(LOG(s(quot(x, s(S(0))))))
log(s(0)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(s(s(x))) → s(LOG(s(QUOT(x, s(s(0))))))
log(s(s(x))) → s(LOG(S(quot(x, s(s(0))))))
quot(0, s(y)) → 0
pred(s(x)) → x
log(s(s(x))) → s(LOG(s(quot(x, S(s(0))))))
s(pred(x)) → pred(s(x))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
log(s(s(x))) → S(LOG(s(quot(x, s(s(0))))))
log(s(s(x))) → s(log(s(QUOT(x, s(S(0))))))

(91) RelADPReductionPairProof (EQUIVALENT transformation)

We use the reduction pair processor [IJCAR24].
The following rules can be oriented strictly (l^# > ann(r))
and therefore we can remove all of its annotations in the right-hand side:
Relative ADPs:


log(s(s(x))) → s(LOG(s(quot(x, s(S(0))))))
log(s(0)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(s(s(x))) → s(LOG(s(QUOT(x, s(s(0))))))
log(s(s(x))) → s(LOG(S(quot(x, s(s(0))))))
quot(0, s(y)) → 0
pred(s(x)) → x
log(s(s(x))) → s(LOG(s(quot(x, S(s(0))))))
s(pred(x)) → pred(s(x))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
log(s(s(x))) → S(LOG(s(quot(x, s(s(0))))))
log(s(s(x))) → s(log(s(QUOT(x, s(S(0))))))


The remaining rules can at least be oriented weakly:
Absolute ADPs:

quot(s(x), s(y)) → s(QUOT(minus(x, y), s(y)))

Relative ADPs:
none


Ordered with Polynomial interpretation [POLO]:

POL(0) = 0   
POL(LOG(x1)) = 3 + x1   
POL(MINUS(x1, x2)) = 0   
POL(PRED(x1)) = 2   
POL(QUOT(x1, x2)) = 0   
POL(S(x1)) = 1 + x1   
POL(log(x1)) = x1   
POL(minus(x1, x2)) = x1   
POL(pred(x1)) = x1   
POL(quot(x1, x2)) = x1   
POL(s(x1)) = 3 + 3·x1   

(92) Obligation:

Relative ADP Problem with
absolute ADPs:

quot(s(x), s(y)) → s(QUOT(minus(x, y), s(y)))

and relative ADPs:

log(s(0)) → 0
s(pred(x)) → pred(s(x))
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
quot(0, s(y)) → 0
pred(s(x)) → x

(93) RelADPCleverAfsProof (SOUND transformation)

We use the first derelatifying processor [IJCAR24].
There are no annotations in relative ADPs, so the relative ADP problem can be transformed into a non-relative DP problem.

Furthermore, We use an argument filter [LPAR04].
Filtering:s_1 = QUOT_2 = log_1 = 0 = pred_1 = minus_2 = 1 quot_2 =
Found this filtering by looking at the following order that orders at least one DP strictly:Combined order from the following AFS and order.
QUOT(x1, x2)  =  QUOT(x1, x2)
s(x1)  =  s(x1)
minus(x1, x2)  =  minus(x1)
pred(x1)  =  x1
0  =  0
quot(x1, x2)  =  quot(x1, x2)

Recursive path order with status [RPO].
Quasi-Precedence:

[0, quot2] > [QUOT2, s1] > minus1

Status:
QUOT2: multiset
s1: multiset
minus1: multiset
0: multiset
quot2: [2,1]

(94) Obligation:

Q DP problem:
The TRS P consists of the following rules:

QUOT0(s0(x), s0(y)) → QUOT0(minus(x), s0(y))

The TRS R consists of the following rules:

s0(pred0(x)) → pred0(s0(x))
log0(s0(00)) → 00
quot0(s0(x), s0(y)) → s0(quot0(minus(x), s0(y)))
minus(x) → pred0(minus(x))
pred0(s0(x)) → s0(pred0(x))
log0(s0(s0(x))) → s0(log0(s0(quot0(x, s0(s0(00))))))
minus(x) → x
quot0(00, s0(y)) → 00
pred0(s0(x)) → x

Q is empty.
We have to consider all (P,Q,R)-chains.

(95) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented rules of the TRS R:

log0(s0(00)) → 00

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0   
POL(QUOT0(x1, x2)) = 2·x1 + x2   
POL(log0(x1)) = 2 + 2·x1   
POL(minus(x1)) = x1   
POL(pred0(x1)) = x1   
POL(quot0(x1, x2)) = x1 + x2   
POL(s0(x1)) = x1   

(96) Obligation:

Q DP problem:
The TRS P consists of the following rules:

QUOT0(s0(x), s0(y)) → QUOT0(minus(x), s0(y))

The TRS R consists of the following rules:

s0(pred0(x)) → pred0(s0(x))
quot0(s0(x), s0(y)) → s0(quot0(minus(x), s0(y)))
minus(x) → pred0(minus(x))
pred0(s0(x)) → s0(pred0(x))
log0(s0(s0(x))) → s0(log0(s0(quot0(x, s0(s0(00))))))
minus(x) → x
quot0(00, s0(y)) → 00
pred0(s0(x)) → x

Q is empty.
We have to consider all (P,Q,R)-chains.

(97) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


QUOT0(s0(x), s0(y)) → QUOT0(minus(x), s0(y))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
POL( s0(x1) ) = x1 + 2

POL( pred0(x1) ) = x1

POL( quot0(x1, x2) ) = x1

POL( minus(x1) ) = x1

POL( log0(x1) ) = max{0, x1 - 2}

POL( 00 ) = 1

POL( QUOT0(x1, x2) ) = 2x1 + 2x2 + 2


The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

s0(pred0(x)) → pred0(s0(x))
quot0(s0(x), s0(y)) → s0(quot0(minus(x), s0(y)))
minus(x) → pred0(minus(x))
pred0(s0(x)) → s0(pred0(x))
log0(s0(s0(x))) → s0(log0(s0(quot0(x, s0(s0(00))))))
minus(x) → x
quot0(00, s0(y)) → 00
pred0(s0(x)) → x

(98) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

s0(pred0(x)) → pred0(s0(x))
quot0(s0(x), s0(y)) → s0(quot0(minus(x), s0(y)))
minus(x) → pred0(minus(x))
pred0(s0(x)) → s0(pred0(x))
log0(s0(s0(x))) → s0(log0(s0(quot0(x, s0(s0(00))))))
minus(x) → x
quot0(00, s0(y)) → 00
pred0(s0(x)) → x

Q is empty.
We have to consider all (P,Q,R)-chains.

(99) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(100) YES

(101) Obligation:

Relative ADP Problem with
absolute ADPs:

quot(s(x), s(y)) → S(quot(minus(x, y), s(y)))

and relative ADPs:

log(s(s(x))) → s(LOG(s(quot(x, s(S(0))))))
log(s(0)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(s(s(x))) → s(LOG(s(QUOT(x, s(s(0))))))
log(s(s(x))) → s(LOG(S(quot(x, s(s(0))))))
quot(0, s(y)) → 0
pred(s(x)) → x
log(s(s(x))) → s(LOG(s(quot(x, S(s(0))))))
s(pred(x)) → pred(s(x))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
log(s(s(x))) → S(LOG(s(quot(x, s(s(0))))))

(102) RelADPReductionPairProof (EQUIVALENT transformation)

We use the reduction pair processor [IJCAR24].
The following rules can be oriented strictly (l^# > ann(r))
and therefore we can remove all of its annotations in the right-hand side:
Absolute ADPs:


quot(s(x), s(y)) → S(quot(minus(x, y), s(y)))

Relative ADPs:

log(s(s(x))) → s(LOG(s(quot(x, s(S(0))))))
log(s(0)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(s(s(x))) → s(LOG(s(QUOT(x, s(s(0))))))
log(s(s(x))) → s(LOG(S(quot(x, s(s(0))))))
quot(0, s(y)) → 0
pred(s(x)) → x
s(pred(x)) → pred(s(x))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
log(s(s(x))) → S(LOG(s(quot(x, s(s(0))))))


The remaining rules can at least be oriented weakly:

Ordered with Polynomial interpretation [POLO]:

POL(0) = 2   
POL(LOG(x1)) = 2 + 3·x1   
POL(MINUS(x1, x2)) = 2·x1   
POL(PRED(x1)) = 0   
POL(QUOT(x1, x2)) = 3·x1   
POL(S(x1)) = 2 + 2·x1   
POL(log(x1)) = 1 + 2·x1   
POL(minus(x1, x2)) = x1   
POL(pred(x1)) = x1   
POL(quot(x1, x2)) = x1   
POL(s(x1)) = 2 + 3·x1   

(103) Obligation:

Relative ADP Problem with
No absolute ADPs, and relative ADPs:

log(s(s(x))) → s(LOG(s(quot(x, S(s(0))))))
log(s(0)) → 0
s(pred(x)) → pred(s(x))
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
quot(0, s(y)) → 0
pred(s(x)) → x

(104) DAbsisEmptyProof (EQUIVALENT transformation)

The relative ADP Problem has an empty P_abs. Hence, no infinite chain exists.

(105) YES

(106) Obligation:

Relative ADP Problem with
absolute ADPs:

quot(s(x), s(y)) → s(QUOT(minus(x, y), s(y)))

and relative ADPs:

log(s(s(x))) → s(LOG(s(quot(x, s(S(0))))))
log(s(0)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(s(s(x))) → s(LOG(s(QUOT(x, s(s(0))))))
log(s(s(x))) → s(LOG(S(quot(x, s(s(0))))))
log(s(s(x))) → s(log(s(QUOT(x, S(s(0))))))
quot(0, s(y)) → 0
pred(s(x)) → x
log(s(s(x))) → s(LOG(s(quot(x, S(s(0))))))
s(pred(x)) → pred(s(x))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
log(s(s(x))) → S(LOG(s(quot(x, s(s(0))))))

(107) RelADPReductionPairProof (EQUIVALENT transformation)

We use the reduction pair processor [IJCAR24].
The following rules can be oriented strictly (l^# > ann(r))
and therefore we can remove all of its annotations in the right-hand side:
Relative ADPs:


log(s(0)) → 0
s(pred(x)) → pred(s(x))
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
log(s(s(x))) → s(log(s(QUOT(x, S(s(0))))))
quot(0, s(y)) → 0
pred(s(x)) → x


The remaining rules can at least be oriented weakly:
Absolute ADPs:

quot(s(x), s(y)) → s(QUOT(minus(x, y), s(y)))

Relative ADPs:

log(s(s(x))) → s(LOG(s(quot(x, s(S(0))))))
log(s(s(x))) → s(LOG(s(quot(x, S(s(0))))))
log(s(s(x))) → s(LOG(s(QUOT(x, s(s(0))))))
log(s(s(x))) → s(LOG(S(quot(x, s(s(0))))))
log(s(s(x))) → S(LOG(s(quot(x, s(s(0))))))


Ordered with Polynomial interpretation [POLO]:

POL(0) = 0   
POL(LOG(x1)) = 1   
POL(MINUS(x1, x2)) = 2 + 2·x1   
POL(PRED(x1)) = 0   
POL(QUOT(x1, x2)) = 0   
POL(S(x1)) = 0   
POL(log(x1)) = 3·x1   
POL(minus(x1, x2)) = x1   
POL(pred(x1)) = x1   
POL(quot(x1, x2)) = x1   
POL(s(x1)) = 1 + 3·x1   

(108) Obligation:

Relative ADP Problem with
absolute ADPs:

quot(s(x), s(y)) → s(QUOT(minus(x, y), s(y)))

and relative ADPs:

log(s(s(x))) → s(LOG(s(quot(x, s(S(0))))))
log(s(0)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(s(s(x))) → s(LOG(s(QUOT(x, s(s(0))))))
log(s(s(x))) → s(LOG(S(quot(x, s(s(0))))))
quot(0, s(y)) → 0
pred(s(x)) → x
log(s(s(x))) → s(LOG(s(quot(x, S(s(0))))))
s(pred(x)) → pred(s(x))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
log(s(s(x))) → S(LOG(s(quot(x, s(s(0))))))

(109) RelADPReductionPairProof (EQUIVALENT transformation)

We use the reduction pair processor [IJCAR24].
The following rules can be oriented strictly (l^# > ann(r))
and therefore we can remove all of its annotations in the right-hand side:
Absolute ADPs:


quot(s(x), s(y)) → s(QUOT(minus(x, y), s(y)))

Relative ADPs:

log(s(s(x))) → s(LOG(s(quot(x, s(S(0))))))
log(s(0)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(s(s(x))) → s(LOG(s(QUOT(x, s(s(0))))))
log(s(s(x))) → s(LOG(S(quot(x, s(s(0))))))
quot(0, s(y)) → 0
pred(s(x)) → x
log(s(s(x))) → s(LOG(s(quot(x, S(s(0))))))
s(pred(x)) → pred(s(x))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
log(s(s(x))) → S(LOG(s(quot(x, s(s(0))))))


No rules with annotations remain.
Ordered with Polynomial interpretation [POLO]:

POL(0) = 0   
POL(LOG(x1)) = 2 + 2·x1   
POL(MINUS(x1, x2)) = 2 + 2·x1 + 2·x1·x2 + 2·x2   
POL(PRED(x1)) = 2   
POL(QUOT(x1, x2)) = 3·x1 + x2   
POL(S(x1)) = 3   
POL(log(x1)) = 2 + 3·x1   
POL(minus(x1, x2)) = x1   
POL(pred(x1)) = x1   
POL(quot(x1, x2)) = x1   
POL(s(x1)) = 3 + 3·x1   

(110) Obligation:

Relative ADP Problem with
No absolute ADPs, and relative ADPs:

log(s(0)) → 0
s(pred(x)) → pred(s(x))
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
quot(0, s(y)) → 0
pred(s(x)) → x

(111) DAbsisEmptyProof (EQUIVALENT transformation)

The relative ADP Problem has an empty P_abs. Hence, no infinite chain exists.

(112) YES

(113) Obligation:

Relative ADP Problem with
absolute ADPs:

quot(s(x), s(y)) → s(quot(minus(x, y), S(y)))

and relative ADPs:

log(s(s(x))) → s(LOG(s(quot(x, s(S(0))))))
log(s(0)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(s(s(x))) → s(LOG(s(QUOT(x, s(s(0))))))
log(s(s(x))) → s(LOG(S(quot(x, s(s(0))))))
log(s(s(x))) → s(log(s(QUOT(x, S(s(0))))))
quot(0, s(y)) → 0
pred(s(x)) → x
log(s(s(x))) → s(LOG(s(quot(x, S(s(0))))))
s(pred(x)) → pred(s(x))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
log(s(s(x))) → S(LOG(s(quot(x, s(s(0))))))

(114) RelADPReductionPairProof (EQUIVALENT transformation)

We use the reduction pair processor [IJCAR24].
The following rules can be oriented strictly (l^# > ann(r))
and therefore we can remove all of its annotations in the right-hand side:
Absolute ADPs:


quot(s(x), s(y)) → s(quot(minus(x, y), S(y)))

Relative ADPs:

log(s(s(x))) → s(LOG(s(quot(x, s(S(0))))))
log(s(0)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(s(s(x))) → s(LOG(s(QUOT(x, s(s(0))))))
log(s(s(x))) → s(LOG(S(quot(x, s(s(0))))))
log(s(s(x))) → s(log(s(QUOT(x, S(s(0))))))
quot(0, s(y)) → 0
pred(s(x)) → x
log(s(s(x))) → s(LOG(s(quot(x, S(s(0))))))
s(pred(x)) → pred(s(x))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x


The remaining rules can at least be oriented weakly:

Ordered with Polynomial interpretation [POLO]:

POL(0) = 0   
POL(LOG(x1)) = 1 + 3·x1   
POL(MINUS(x1, x2)) = 2 + 2·x1   
POL(PRED(x1)) = 2   
POL(QUOT(x1, x2)) = 2 + 2·x2   
POL(S(x1)) = 3 + 2·x1   
POL(log(x1)) = 3 + 3·x1   
POL(minus(x1, x2)) = x1   
POL(pred(x1)) = x1   
POL(quot(x1, x2)) = x1   
POL(s(x1)) = 3 + 3·x1   

(115) Obligation:

Relative ADP Problem with
No absolute ADPs, and relative ADPs:

log(s(0)) → 0
s(pred(x)) → pred(s(x))
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
log(s(s(x))) → S(LOG(s(quot(x, s(s(0))))))
quot(0, s(y)) → 0
pred(s(x)) → x

(116) DAbsisEmptyProof (EQUIVALENT transformation)

The relative ADP Problem has an empty P_abs. Hence, no infinite chain exists.

(117) YES

(118) Obligation:

Relative ADP Problem with
absolute ADPs:

quot(0, s(y)) → 0

and relative ADPs:

log(s(s(x))) → s(LOG(s(quot(x, s(S(0))))))
log(s(0)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(s(s(x))) → s(LOG(s(QUOT(x, s(s(0))))))
log(s(s(x))) → s(LOG(S(quot(x, s(s(0))))))
pred(s(x)) → x
log(s(s(x))) → s(LOG(s(quot(x, S(s(0))))))
s(pred(x)) → pred(s(x))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
log(s(s(x))) → S(LOG(s(quot(x, s(s(0))))))
log(s(s(x))) → s(log(s(QUOT(x, s(S(0))))))

(119) RelADPReductionPairProof (EQUIVALENT transformation)

We use the reduction pair processor [IJCAR24].
The following rules can be oriented strictly (l^# > ann(r))
and therefore we can remove all of its annotations in the right-hand side:
Absolute ADPs:


quot(0, s(y)) → 0

Relative ADPs:

log(s(0)) → 0
s(pred(x)) → pred(s(x))
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
pred(s(x)) → x


The remaining rules can at least be oriented weakly:

Ordered with Polynomial interpretation [POLO]:

POL(0) = 1   
POL(LOG(x1)) = 2·x1   
POL(MINUS(x1, x2)) = 2 + 2·x1 + 2·x2   
POL(PRED(x1)) = x12   
POL(QUOT(x1, x2)) = 2·x1   
POL(S(x1)) = 0   
POL(log(x1)) = x1   
POL(minus(x1, x2)) = x1   
POL(pred(x1)) = x1   
POL(quot(x1, x2)) = x1   
POL(s(x1)) = 3·x1   

(120) Obligation:

Relative ADP Problem with
No absolute ADPs, and relative ADPs:

log(s(s(x))) → s(LOG(s(quot(x, s(S(0))))))
log(s(0)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(s(s(x))) → s(LOG(s(QUOT(x, s(s(0))))))
log(s(s(x))) → s(LOG(S(quot(x, s(s(0))))))
quot(0, s(y)) → 0
pred(s(x)) → x
log(s(s(x))) → s(LOG(s(quot(x, S(s(0))))))
s(pred(x)) → pred(s(x))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
log(s(s(x))) → S(LOG(s(quot(x, s(s(0))))))
log(s(s(x))) → s(log(s(QUOT(x, s(S(0))))))

(121) DAbsisEmptyProof (EQUIVALENT transformation)

The relative ADP Problem has an empty P_abs. Hence, no infinite chain exists.

(122) YES

(123) Obligation:

Relative ADP Problem with
absolute ADPs:

quot(s(x), s(y)) → s(quot(MINUS(x, y), s(y)))

and relative ADPs:

log(s(s(x))) → s(LOG(s(quot(x, s(S(0))))))
log(s(0)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(s(s(x))) → s(LOG(s(QUOT(x, s(s(0))))))
log(s(s(x))) → s(LOG(S(quot(x, s(s(0))))))
quot(0, s(y)) → 0
pred(s(x)) → x
log(s(s(x))) → s(LOG(s(quot(x, S(s(0))))))
s(pred(x)) → pred(s(x))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
log(s(s(x))) → s(log(S(QUOT(x, s(s(0))))))
minus(x, 0) → x
log(s(s(x))) → S(LOG(s(quot(x, s(s(0))))))

(124) RelADPReductionPairProof (EQUIVALENT transformation)

We use the reduction pair processor [IJCAR24].
The following rules can be oriented strictly (l^# > ann(r))
and therefore we can remove all of its annotations in the right-hand side:
Relative ADPs:


log(s(s(x))) → s(LOG(s(quot(x, s(S(0))))))
log(s(0)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(s(s(x))) → s(LOG(s(QUOT(x, s(s(0))))))
log(s(s(x))) → s(LOG(S(quot(x, s(s(0))))))
quot(0, s(y)) → 0
pred(s(x)) → x
log(s(s(x))) → s(LOG(s(quot(x, S(s(0))))))
s(pred(x)) → pred(s(x))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
log(s(s(x))) → s(log(S(QUOT(x, s(s(0))))))
minus(x, 0) → x
log(s(s(x))) → S(LOG(s(quot(x, s(s(0))))))


The remaining rules can at least be oriented weakly:
Absolute ADPs:

quot(s(x), s(y)) → s(quot(MINUS(x, y), s(y)))

Relative ADPs:
none


Ordered with Polynomial interpretation [POLO]:

POL(0) = 3   
POL(LOG(x1)) = 3 + 3·x1   
POL(MINUS(x1, x2)) = 3 + 2·x1   
POL(PRED(x1)) = 0   
POL(QUOT(x1, x2)) = x1   
POL(S(x1)) = x1   
POL(log(x1)) = 2 + 2·x1   
POL(minus(x1, x2)) = x1   
POL(pred(x1)) = x1   
POL(quot(x1, x2)) = x1   
POL(s(x1)) = 3 + 2·x1   

(125) Obligation:

Relative ADP Problem with
absolute ADPs:

quot(s(x), s(y)) → s(quot(MINUS(x, y), s(y)))

and relative ADPs:

log(s(0)) → 0
s(pred(x)) → pred(s(x))
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
quot(0, s(y)) → 0
pred(s(x)) → x

(126) RelADPDepGraphProof (EQUIVALENT transformation)

We use the relative dependency graph processor [IJCAR24].
The approximation of the Relative Dependency Graph contains:
0 SCCs with nodes from P_abs,
0 Lassos,
Result: This relative DT problem is equivalent to 0 subproblems.

(127) TRUE

(128) Obligation:

Relative ADP Problem with
absolute ADPs:

quot(s(x), s(y)) → s(QUOT(minus(x, y), s(y)))

and relative ADPs:

log(s(s(x))) → s(LOG(s(quot(x, s(S(0))))))
log(s(0)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(s(s(x))) → s(LOG(s(QUOT(x, s(s(0))))))
log(s(s(x))) → s(LOG(S(quot(x, s(s(0))))))
quot(0, s(y)) → 0
pred(s(x)) → x
log(s(s(x))) → s(LOG(s(quot(x, S(s(0))))))
s(pred(x)) → pred(s(x))
log(s(s(x))) → S(log(s(QUOT(x, s(s(0))))))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
log(s(s(x))) → S(LOG(s(quot(x, s(s(0))))))

(129) RelADPReductionPairProof (EQUIVALENT transformation)

We use the reduction pair processor [IJCAR24].
The following rules can be oriented strictly (l^# > ann(r))
and therefore we can remove all of its annotations in the right-hand side:
Absolute ADPs:


quot(s(x), s(y)) → s(QUOT(minus(x, y), s(y)))

Relative ADPs:

log(s(s(x))) → s(LOG(s(quot(x, s(S(0))))))
log(s(0)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(s(s(x))) → s(LOG(s(QUOT(x, s(s(0))))))
log(s(s(x))) → s(LOG(S(quot(x, s(s(0))))))
quot(0, s(y)) → 0
pred(s(x)) → x
log(s(s(x))) → s(LOG(s(quot(x, S(s(0))))))
s(pred(x)) → pred(s(x))
log(s(s(x))) → S(log(s(QUOT(x, s(s(0))))))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
log(s(s(x))) → S(LOG(s(quot(x, s(s(0))))))


No rules with annotations remain.
Ordered with Polynomial interpretation [POLO]:

POL(0) = 0   
POL(LOG(x1)) = 2 + 2·x1   
POL(MINUS(x1, x2)) = 0   
POL(PRED(x1)) = 2   
POL(QUOT(x1, x2)) = 1 + 2·x1 + x2   
POL(S(x1)) = 1 + x1   
POL(log(x1)) = x1   
POL(minus(x1, x2)) = x1   
POL(pred(x1)) = x1   
POL(quot(x1, x2)) = x1   
POL(s(x1)) = 3 + 3·x1   

(130) Obligation:

Relative ADP Problem with
No absolute ADPs, and relative ADPs:

log(s(0)) → 0
s(pred(x)) → pred(s(x))
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
quot(0, s(y)) → 0
pred(s(x)) → x

(131) DAbsisEmptyProof (EQUIVALENT transformation)

The relative ADP Problem has an empty P_abs. Hence, no infinite chain exists.

(132) YES

(133) Obligation:

Relative ADP Problem with
absolute ADPs:

quot(0, s(y)) → 0

and relative ADPs:

log(s(s(x))) → s(LOG(s(quot(x, s(S(0))))))
log(s(0)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(s(s(x))) → s(LOG(s(QUOT(x, s(s(0))))))
log(s(s(x))) → s(LOG(S(quot(x, s(s(0))))))
pred(s(x)) → x
log(s(s(x))) → s(LOG(s(quot(x, S(s(0))))))
s(pred(x)) → pred(s(x))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
log(s(s(x))) → s(log(S(QUOT(x, s(s(0))))))
minus(x, 0) → x
log(s(s(x))) → S(LOG(s(quot(x, s(s(0))))))

(134) RelADPReductionPairProof (EQUIVALENT transformation)

We use the reduction pair processor [IJCAR24].
The following rules can be oriented strictly (l^# > ann(r))
and therefore we can remove all of its annotations in the right-hand side:
Absolute ADPs:


quot(0, s(y)) → 0

Relative ADPs:

log(s(s(x))) → s(LOG(s(quot(x, s(S(0))))))
log(s(0)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(s(s(x))) → s(LOG(s(QUOT(x, s(s(0))))))
log(s(s(x))) → s(LOG(S(quot(x, s(s(0))))))
pred(s(x)) → x
log(s(s(x))) → s(LOG(s(quot(x, S(s(0))))))
s(pred(x)) → pred(s(x))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
log(s(s(x))) → s(log(S(QUOT(x, s(s(0))))))
minus(x, 0) → x
log(s(s(x))) → S(LOG(s(quot(x, s(s(0))))))


No rules with annotations remain.
Ordered with Polynomial interpretation [POLO]:

POL(0) = 1   
POL(LOG(x1)) = 2 + 3·x1   
POL(MINUS(x1, x2)) = 2 + 2·x2   
POL(PRED(x1)) = 2 + 3·x12   
POL(QUOT(x1, x2)) = 3   
POL(S(x1)) = x1   
POL(log(x1)) = 2 + 3·x1   
POL(minus(x1, x2)) = x1   
POL(pred(x1)) = x1   
POL(quot(x1, x2)) = x1   
POL(s(x1)) = 2 + 3·x1   

(135) Obligation:

Relative ADP Problem with
No absolute ADPs, and relative ADPs:

log(s(0)) → 0
s(pred(x)) → pred(s(x))
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
quot(0, s(y)) → 0
pred(s(x)) → x

(136) DAbsisEmptyProof (EQUIVALENT transformation)

The relative ADP Problem has an empty P_abs. Hence, no infinite chain exists.

(137) YES

(138) Obligation:

Relative ADP Problem with
absolute ADPs:

quot(s(x), s(y)) → S(quot(minus(x, y), s(y)))

and relative ADPs:

log(s(s(x))) → s(LOG(s(quot(x, s(S(0))))))
log(s(0)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(s(s(x))) → s(LOG(s(QUOT(x, s(s(0))))))
log(s(s(x))) → s(LOG(S(quot(x, s(s(0))))))
quot(0, s(y)) → 0
pred(s(x)) → x
log(s(s(x))) → s(LOG(s(quot(x, S(s(0))))))
s(pred(x)) → pred(s(x))
log(s(s(x))) → S(log(s(QUOT(x, s(s(0))))))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
log(s(s(x))) → S(LOG(s(quot(x, s(s(0))))))

(139) RelADPReductionPairProof (EQUIVALENT transformation)

We use the reduction pair processor [IJCAR24].
The following rules can be oriented strictly (l^# > ann(r))
and therefore we can remove all of its annotations in the right-hand side:
Absolute ADPs:


quot(s(x), s(y)) → S(quot(minus(x, y), s(y)))

Relative ADPs:

log(s(s(x))) → s(LOG(s(quot(x, s(S(0))))))
log(s(0)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(s(s(x))) → s(LOG(s(QUOT(x, s(s(0))))))
log(s(s(x))) → s(LOG(S(quot(x, s(s(0))))))
quot(0, s(y)) → 0
pred(s(x)) → x
log(s(s(x))) → s(LOG(s(quot(x, S(s(0))))))
s(pred(x)) → pred(s(x))
log(s(s(x))) → S(log(s(QUOT(x, s(s(0))))))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
log(s(s(x))) → S(LOG(s(quot(x, s(s(0))))))


No rules with annotations remain.
Ordered with Polynomial interpretation [POLO]:

POL(0) = 1   
POL(LOG(x1)) = 3·x1   
POL(MINUS(x1, x2)) = 2   
POL(PRED(x1)) = 0   
POL(QUOT(x1, x2)) = 1 + x2   
POL(S(x1)) = 3   
POL(log(x1)) = x1   
POL(minus(x1, x2)) = x1   
POL(pred(x1)) = x1   
POL(quot(x1, x2)) = x1   
POL(s(x1)) = 3 + 3·x1   

(140) Obligation:

Relative ADP Problem with
No absolute ADPs, and relative ADPs:

log(s(0)) → 0
s(pred(x)) → pred(s(x))
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
quot(0, s(y)) → 0
pred(s(x)) → x

(141) DAbsisEmptyProof (EQUIVALENT transformation)

The relative ADP Problem has an empty P_abs. Hence, no infinite chain exists.

(142) YES

(143) Obligation:

Relative ADP Problem with
absolute ADPs:

quot(s(x), s(y)) → s(quot(MINUS(x, y), s(y)))

and relative ADPs:

log(s(s(x))) → s(LOG(s(quot(x, s(S(0))))))
log(s(0)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(s(s(x))) → s(LOG(s(QUOT(x, s(s(0))))))
log(s(s(x))) → s(LOG(S(quot(x, s(s(0))))))
quot(0, s(y)) → 0
pred(s(x)) → x
log(s(s(x))) → s(LOG(s(quot(x, S(s(0))))))
s(pred(x)) → pred(s(x))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
log(s(s(x))) → S(LOG(s(quot(x, s(s(0))))))
log(s(s(x))) → s(log(s(QUOT(x, s(S(0))))))

(144) RelADPReductionPairProof (EQUIVALENT transformation)

We use the reduction pair processor [IJCAR24].
The following rules can be oriented strictly (l^# > ann(r))
and therefore we can remove all of its annotations in the right-hand side:
Absolute ADPs:


quot(s(x), s(y)) → s(quot(MINUS(x, y), s(y)))

Relative ADPs:

log(s(s(x))) → s(LOG(s(quot(x, s(S(0))))))
log(s(0)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(s(s(x))) → s(LOG(S(quot(x, s(s(0))))))
quot(0, s(y)) → 0
pred(s(x)) → x
log(s(s(x))) → s(LOG(s(quot(x, S(s(0))))))
s(pred(x)) → pred(s(x))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
log(s(s(x))) → S(LOG(s(quot(x, s(s(0))))))
log(s(s(x))) → s(log(s(QUOT(x, s(S(0))))))


The remaining rules can at least be oriented weakly:

Ordered with Polynomial interpretation [POLO]:

POL(0) = 1   
POL(LOG(x1)) = 3·x1   
POL(MINUS(x1, x2)) = 3 + x2   
POL(PRED(x1)) = 2 + x1   
POL(QUOT(x1, x2)) = 2 + x2   
POL(S(x1)) = 3   
POL(log(x1)) = 3 + 2·x1   
POL(minus(x1, x2)) = x1   
POL(pred(x1)) = x1   
POL(quot(x1, x2)) = x1   
POL(s(x1)) = 3 + x1   

(145) Obligation:

Relative ADP Problem with
No absolute ADPs, and relative ADPs:

log(s(0)) → 0
s(pred(x)) → pred(s(x))
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(s(s(x))) → s(LOG(s(QUOT(x, s(s(0))))))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
quot(0, s(y)) → 0
pred(s(x)) → x

(146) DAbsisEmptyProof (EQUIVALENT transformation)

The relative ADP Problem has an empty P_abs. Hence, no infinite chain exists.

(147) YES

(148) Obligation:

Relative ADP Problem with
absolute ADPs:

quot(s(x), s(y)) → s(quot(minus(x, y), S(y)))

and relative ADPs:

log(s(s(x))) → s(LOG(s(quot(x, s(S(0))))))
log(s(0)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(s(s(x))) → s(LOG(s(QUOT(x, s(s(0))))))
log(s(s(x))) → s(LOG(S(quot(x, s(s(0))))))
quot(0, s(y)) → 0
pred(s(x)) → x
log(s(s(x))) → s(LOG(s(quot(x, S(s(0))))))
s(pred(x)) → pred(s(x))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
log(s(s(x))) → S(LOG(s(quot(x, s(s(0))))))
log(s(s(x))) → s(log(s(QUOT(x, s(S(0))))))

(149) RelADPReductionPairProof (EQUIVALENT transformation)

We use the reduction pair processor [IJCAR24].
The following rules can be oriented strictly (l^# > ann(r))
and therefore we can remove all of its annotations in the right-hand side:
Relative ADPs:


log(s(0)) → 0
s(pred(x)) → pred(s(x))
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
log(s(s(x))) → s(log(s(QUOT(x, s(S(0))))))
quot(0, s(y)) → 0
pred(s(x)) → x


The remaining rules can at least be oriented weakly:
Absolute ADPs:

quot(s(x), s(y)) → s(quot(minus(x, y), S(y)))

Relative ADPs:

log(s(s(x))) → s(LOG(s(quot(x, s(S(0))))))
log(s(s(x))) → s(LOG(s(quot(x, S(s(0))))))
log(s(s(x))) → s(LOG(s(QUOT(x, s(s(0))))))
log(s(s(x))) → s(LOG(S(quot(x, s(s(0))))))
log(s(s(x))) → S(LOG(s(quot(x, s(s(0))))))


Ordered with Polynomial interpretation [POLO]:

POL(0) = 0   
POL(LOG(x1)) = 2 + 3·x1   
POL(MINUS(x1, x2)) = 2 + 2·x1 + 2·x2   
POL(PRED(x1)) = x12   
POL(QUOT(x1, x2)) = 2·x1 + x2   
POL(S(x1)) = 0   
POL(log(x1)) = 0   
POL(minus(x1, x2)) = x1   
POL(pred(x1)) = x1   
POL(quot(x1, x2)) = 2·x1   
POL(s(x1)) = 3·x1   

(150) Obligation:

Relative ADP Problem with
absolute ADPs:

quot(s(x), s(y)) → s(quot(minus(x, y), S(y)))

and relative ADPs:

log(s(s(x))) → s(LOG(s(quot(x, s(S(0))))))
log(s(0)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(s(s(x))) → s(LOG(s(QUOT(x, s(s(0))))))
log(s(s(x))) → s(LOG(S(quot(x, s(s(0))))))
quot(0, s(y)) → 0
pred(s(x)) → x
log(s(s(x))) → s(LOG(s(quot(x, S(s(0))))))
s(pred(x)) → pred(s(x))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
log(s(s(x))) → S(LOG(s(quot(x, s(s(0))))))

(151) RelADPReductionPairProof (EQUIVALENT transformation)

We use the reduction pair processor [IJCAR24].
The following rules can be oriented strictly (l^# > ann(r))
and therefore we can remove all of its annotations in the right-hand side:
Absolute ADPs:


quot(s(x), s(y)) → s(quot(minus(x, y), S(y)))

Relative ADPs:

log(s(s(x))) → s(LOG(s(quot(x, s(S(0))))))
log(s(0)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(s(s(x))) → s(LOG(s(QUOT(x, s(s(0))))))
log(s(s(x))) → s(LOG(S(quot(x, s(s(0))))))
quot(0, s(y)) → 0
pred(s(x)) → x
log(s(s(x))) → s(LOG(s(quot(x, S(s(0))))))
s(pred(x)) → pred(s(x))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
log(s(s(x))) → S(LOG(s(quot(x, s(s(0))))))


No rules with annotations remain.
Ordered with Polynomial interpretation [POLO]:

POL(0) = 0   
POL(LOG(x1)) = 2 + 2·x1   
POL(MINUS(x1, x2)) = 2 + 2·x1 + 2·x2   
POL(PRED(x1)) = 2   
POL(QUOT(x1, x2)) = 2 + x2   
POL(S(x1)) = 2 + 2·x1   
POL(log(x1)) = x1   
POL(minus(x1, x2)) = x1   
POL(pred(x1)) = x1   
POL(quot(x1, x2)) = x1   
POL(s(x1)) = 3 + 3·x1   

(152) Obligation:

Relative ADP Problem with
No absolute ADPs, and relative ADPs:

log(s(0)) → 0
s(pred(x)) → pred(s(x))
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
quot(0, s(y)) → 0
pred(s(x)) → x

(153) DAbsisEmptyProof (EQUIVALENT transformation)

The relative ADP Problem has an empty P_abs. Hence, no infinite chain exists.

(154) YES

(155) Obligation:

Relative ADP Problem with
absolute ADPs:

quot(s(x), s(y)) → s(QUOT(minus(x, y), s(y)))

and relative ADPs:

log(s(s(x))) → s(LOG(s(quot(x, s(S(0))))))
log(s(0)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(s(s(x))) → s(LOG(s(QUOT(x, s(s(0))))))
log(s(s(x))) → s(LOG(S(quot(x, s(s(0))))))
quot(0, s(y)) → 0
pred(s(x)) → x
log(s(s(x))) → s(LOG(s(quot(x, S(s(0))))))
s(pred(x)) → pred(s(x))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
log(s(s(x))) → S(LOG(s(quot(x, s(s(0))))))

(156) RelADPReductionPairProof (EQUIVALENT transformation)

We use the reduction pair processor [IJCAR24].
The following rules can be oriented strictly (l^# > ann(r))
and therefore we can remove all of its annotations in the right-hand side:
Absolute ADPs:


quot(s(x), s(y)) → s(QUOT(minus(x, y), s(y)))

Relative ADPs:

log(s(s(x))) → s(LOG(s(quot(x, s(S(0))))))
log(s(0)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(s(s(x))) → s(LOG(s(QUOT(x, s(s(0))))))
log(s(s(x))) → s(LOG(S(quot(x, s(s(0))))))
quot(0, s(y)) → 0
pred(s(x)) → x
log(s(s(x))) → s(LOG(s(quot(x, S(s(0))))))
s(pred(x)) → pred(s(x))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
log(s(s(x))) → S(LOG(s(quot(x, s(s(0))))))


No rules with annotations remain.
Ordered with Polynomial interpretation [POLO]:

POL(0) = 0   
POL(LOG(x1)) = 2 + 2·x1   
POL(MINUS(x1, x2)) = 2 + 2·x1 + 2·x1·x2 + 2·x2   
POL(PRED(x1)) = 2   
POL(QUOT(x1, x2)) = 3·x1 + x2   
POL(S(x1)) = 3   
POL(log(x1)) = 2 + 3·x1   
POL(minus(x1, x2)) = x1   
POL(pred(x1)) = x1   
POL(quot(x1, x2)) = x1   
POL(s(x1)) = 3 + 3·x1   

(157) Obligation:

Relative ADP Problem with
No absolute ADPs, and relative ADPs:

log(s(0)) → 0
s(pred(x)) → pred(s(x))
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(x, s(y)) → pred(minus(x, y))
pred(s(x)) → s(pred(x))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
quot(0, s(y)) → 0
pred(s(x)) → x

(158) DAbsisEmptyProof (EQUIVALENT transformation)

The relative ADP Problem has an empty P_abs. Hence, no infinite chain exists.

(159) YES