YES Termination proof of INVY_15_3.8b_rand.ari

Termination of the given RelTRS could be proven:

0 RelTRS
↳1 RelTRStoRelADPProof (⇔, 0 ms)
↳2 RelADPP
↳3 RelADPDepGraphProof (⇔, 0 ms)
↳4 AND
    ↳5 RelADPP
        ↳6 RelADPCleverAfsProof (⇒, 130 ms)
        ↳7 QDP
        ↳8 MRRProof (⇔, 0 ms)
        ↳9 QDP
        ↳10 MRRProof (⇔, 10 ms)
        ↳11 QDP
        ↳12 QDPOrderProof (⇔, 6 ms)
        ↳13 QDP
        ↳14 PisEmptyProof (⇔, 0 ms)
        ↳15 YES
    ↳16 RelADPP
        ↳17 RelADPCleverAfsProof (⇒, 128 ms)
        ↳18 QDP
        ↳19 MRRProof (⇔, 0 ms)
        ↳20 QDP
        ↳21 MRRProof (⇔, 9 ms)
        ↳22 QDP
        ↳23 QDPOrderProof (⇔, 0 ms)
        ↳24 QDP
        ↳25 DependencyGraphProof (⇔, 0 ms)
        ↳26 TRUE
    ↳27 RelADPP
        ↳28 RelADPCleverAfsProof (⇒, 105 ms)
        ↳29 QDP
        ↳30 MRRProof (⇔, 0 ms)
        ↳31 QDP
        ↳32 MRRProof (⇔, 11 ms)
        ↳33 QDP
        ↳34 MRRProof (⇔, 0 ms)
        ↳35 QDP
        ↳36 QDPOrderProof (⇔, 35 ms)
        ↳37 QDP
        ↳38 PisEmptyProof (⇔, 0 ms)
        ↳39 YES
    ↳40 RelADPP
        ↳41 RelADPCleverAfsProof (⇒, 108 ms)
        ↳42 QDP
        ↳43 MRRProof (⇔, 0 ms)
        ↳44 QDP
        ↳45 MRRProof (⇔, 0 ms)
        ↳46 QDP
        ↳47 MRRProof (⇔, 1 ms)
        ↳48 QDP
        ↳49 QDPOrderProof (⇔, 0 ms)
        ↳50 QDP
        ↳51 PisEmptyProof (⇔, 0 ms)
        ↳52 YES

(0) Obligation:

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

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(0, y) → 0
minus(s(x), y) → if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y) → 0
if_minus(false, s(x), y) → s(minus(x, y))
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
log(s(0)) → 0
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))

The relative TRS consists of the following S rules:

rand(x) → rand(s(x))
rand(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:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → LE(x, y)
minus(0, y) → 0
minus(s(x), y) → IF_MINUS(le(s(x), y), s(x), y)
minus(s(x), y) → if_minus(LE(s(x), y), s(x), y)
if_minus(true, s(x), y) → 0
if_minus(false, s(x), y) → s(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)))
log(s(0)) → 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))))))

and relative ADPs:

rand(x) → RAND(s(x))
rand(x) → x

(3) RelADPDepGraphProof (EQUIVALENT transformation)

We use the relative dependency graph processor [IJCAR24].
The approximation of the Relative Dependency Graph contains:
4 SCCs with nodes from P_abs,
0 Lassos,
Result: This relative DT problem is equivalent to 4 subproblems.

(4) Complex Obligation (AND)

(5) Obligation:

Relative ADP Problem with
absolute ADPs:

le(s(x), s(y)) → LE(x, y)

and relative ADPs:

minus(s(x), y) → if_minus(le(s(x), y), s(x), y)
log(s(0)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
if_minus(false, s(x), y) → s(minus(x, y))
le(s(x), 0) → false
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
le(0, y) → true
rand(x) → rand(s(x))
if_minus(true, s(x), y) → 0
rand(x) → x
minus(0, y) → 0
quot(0, s(y)) → 0

(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 = true = LE_2 = 0 log_1 = le_2 = 0, 1 0 = minus_2 = 1 if_minus_3 = 0, 2 rand_1 = quot_2 = 1 false =
Found this filtering by looking at the following order that orders at least one DP strictly:Combined order from the following AFS and order.
LE(x1, x2)  =  LE(x2)
s(x1)  =  s(x1)
quot(x1, x2)  =  x1
minus(x1, x2)  =  x1
if_minus(x1, x2, x3)  =  x2
le(x1, x2)  =  le
0  =  0
false  =  false
true  =  true

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

[le, true] > false > [s1, 0]

Status:
LE1: [1]
s1: multiset
le: []
0: multiset
false: multiset
true: multiset

(7) Obligation:

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

LE(s0(y)) → LE(y)

The TRS R consists of the following rules:

minus(s0(x)) → if_minus(s0(x))
log0(s0(00)) → 00
quot(s0(x)) → s0(quot(minus(x)))
le → le
le → false0
rand0(x) → rand0(s0(x))
if_minus(s0(x)) → 00
minus(00) → 00
quot(00) → 00
if_minus(s0(x)) → s0(minus(x))
log0(s0(s0(x))) → s0(log0(s0(quot(x))))
le → true0
rand0(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
le → false0
rand0(x) → x

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0   
POL(LE(x1)) = x1   
POL(false0) = 0   
POL(if_minus(x1)) = x1   
POL(le) = 2   
POL(log0(x1)) = 1 + 2·x1   
POL(minus(x1)) = x1   
POL(quot(x1)) = x1   
POL(rand0(x1)) = 2 + x1   
POL(s0(x1)) = x1   
POL(true0) = 2   

(9) Obligation:

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

LE(s0(y)) → LE(y)

The TRS R consists of the following rules:

minus(s0(x)) → if_minus(s0(x))
quot(s0(x)) → s0(quot(minus(x)))
le → le
rand0(x) → rand0(s0(x))
if_minus(s0(x)) → 00
minus(00) → 00
quot(00) → 00
if_minus(s0(x)) → s0(minus(x))
log0(s0(s0(x))) → s0(log0(s0(quot(x))))
le → true0

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 rules of the TRS R:

le → true0

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0   
POL(LE(x1)) = x1   
POL(if_minus(x1)) = x1   
POL(le) = 2   
POL(log0(x1)) = x1   
POL(minus(x1)) = x1   
POL(quot(x1)) = x1   
POL(rand0(x1)) = x1   
POL(s0(x1)) = x1   
POL(true0) = 1   

(11) Obligation:

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

LE(s0(y)) → LE(y)

The TRS R consists of the following rules:

minus(s0(x)) → if_minus(s0(x))
quot(s0(x)) → s0(quot(minus(x)))
le → le
rand0(x) → rand0(s0(x))
if_minus(s0(x)) → 00
minus(00) → 00
quot(00) → 00
if_minus(s0(x)) → s0(minus(x))
log0(s0(s0(x))) → s0(log0(s0(quot(x))))

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

(12) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


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

POL( s0(x1) ) = 2x1 + 2

POL( if_minus(x1) ) = x1

POL( quot(x1) ) = x1

POL( le ) = 0

POL( rand0(x1) ) = 2

POL( 00 ) = 0

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

POL( LE(x1) ) = 2x1


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

minus(s0(x)) → if_minus(s0(x))
quot(s0(x)) → s0(quot(minus(x)))
le → le
rand0(x) → rand0(s0(x))
if_minus(s0(x)) → 00
minus(00) → 00
quot(00) → 00
if_minus(s0(x)) → s0(minus(x))
log0(s0(s0(x))) → s0(log0(s0(quot(x))))

(13) Obligation:

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

minus(s0(x)) → if_minus(s0(x))
quot(s0(x)) → s0(quot(minus(x)))
le → le
rand0(x) → rand0(s0(x))
if_minus(s0(x)) → 00
minus(00) → 00
quot(00) → 00
if_minus(s0(x)) → s0(minus(x))
log0(s0(s0(x))) → s0(log0(s0(quot(x))))

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

(14) PisEmptyProof (EQUIVALENT transformation)

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

(15) YES

(16) Obligation:

Relative ADP Problem with
absolute ADPs:

if_minus(false, s(x), y) → s(MINUS(x, y))
minus(s(x), y) → IF_MINUS(le(s(x), y), s(x), y)

and relative ADPs:

minus(s(x), y) → if_minus(le(s(x), y), s(x), y)
log(s(0)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
le(s(x), s(y)) → le(x, y)
le(s(x), 0) → false
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
le(0, y) → true
rand(x) → rand(s(x))
if_minus(true, s(x), y) → 0
rand(x) → x
minus(0, y) → 0
quot(0, s(y)) → 0

(17) 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 = true = log_1 = le_2 = 0, 1 0 = minus_2 = 1 IF_MINUS_3 = 0 if_minus_3 = 0, 2 rand_1 = false = 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.
IF_MINUS(x1, x2, x3)  =  IF_MINUS(x2, x3)
false  =  false
s(x1)  =  s(x1)
MINUS(x1, x2)  =  MINUS(x1, x2)
le(x1, x2)  =  le
0  =  0
true  =  true
quot(x1, x2)  =  x1
minus(x1, x2)  =  x1
if_minus(x1, x2, x3)  =  x2

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

s1 > le > false > [IFMINUS2, MINUS2] > true
s1 > 0 > true

Status:
IFMINUS2: [2,1]
false: multiset
s1: [1]
MINUS2: [2,1]
le: multiset
0: multiset
true: multiset

(18) Obligation:

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

IF_MINUS(s0(x), y) → MINUS0(x, y)
MINUS0(s0(x), y) → IF_MINUS(s0(x), y)

The TRS R consists of the following rules:

minus(s0(x)) → if_minus(s0(x))
log0(s0(00)) → 00
quot(s0(x)) → s0(quot(minus(x)))
le → le
le → false0
rand0(x) → rand0(s0(x))
if_minus(s0(x)) → 00
minus(00) → 00
quot(00) → 00
if_minus(s0(x)) → s0(minus(x))
log0(s0(s0(x))) → s0(log0(s0(quot(x))))
le → true0
rand0(x) → x

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

(19) 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
le → true0
rand0(x) → x

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0   
POL(IF_MINUS(x1, x2)) = x1 + x2   
POL(MINUS0(x1, x2)) = x1 + x2   
POL(false0) = 2   
POL(if_minus(x1)) = x1   
POL(le) = 2   
POL(log0(x1)) = 1 + x1   
POL(minus(x1)) = x1   
POL(quot(x1)) = x1   
POL(rand0(x1)) = 2 + x1   
POL(s0(x1)) = x1   
POL(true0) = 0   

(20) Obligation:

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

IF_MINUS(s0(x), y) → MINUS0(x, y)
MINUS0(s0(x), y) → IF_MINUS(s0(x), y)

The TRS R consists of the following rules:

minus(s0(x)) → if_minus(s0(x))
quot(s0(x)) → s0(quot(minus(x)))
le → le
le → false0
rand0(x) → rand0(s0(x))
if_minus(s0(x)) → 00
minus(00) → 00
quot(00) → 00
if_minus(s0(x)) → s0(minus(x))
log0(s0(s0(x))) → s0(log0(s0(quot(x))))

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

(21) 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:

le → false0

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0   
POL(IF_MINUS(x1, x2)) = x1 + x2   
POL(MINUS0(x1, x2)) = x1 + x2   
POL(false0) = 1   
POL(if_minus(x1)) = x1   
POL(le) = 2   
POL(log0(x1)) = x1   
POL(minus(x1)) = x1   
POL(quot(x1)) = x1   
POL(rand0(x1)) = x1   
POL(s0(x1)) = x1   

(22) Obligation:

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

IF_MINUS(s0(x), y) → MINUS0(x, y)
MINUS0(s0(x), y) → IF_MINUS(s0(x), y)

The TRS R consists of the following rules:

minus(s0(x)) → if_minus(s0(x))
quot(s0(x)) → s0(quot(minus(x)))
le → le
rand0(x) → rand0(s0(x))
if_minus(s0(x)) → 00
minus(00) → 00
quot(00) → 00
if_minus(s0(x)) → s0(minus(x))
log0(s0(s0(x))) → s0(log0(s0(quot(x))))

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

(23) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


MINUS0(s0(x), y) → IF_MINUS(s0(x), y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
IF_MINUS(x1, x2)  =  x1
s0(x1)  =  s0(x1)
MINUS0(x1, x2)  =  MINUS0(x1)
minus(x1)  =  x1
if_minus(x1)  =  x1
quot(x1)  =  x1
le  =  le
rand0(x1)  =  rand0
00  =  00
log0(x1)  =  log0(x1)

Recursive path order with status [RPO].
Quasi-Precedence:
rand0 > [s01, MINUS01, 00]
log01 > [s01, MINUS01, 00]

Status:
s01: multiset
MINUS01: multiset
le: multiset
rand0: multiset
00: multiset
log01: multiset


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

minus(s0(x)) → if_minus(s0(x))
quot(s0(x)) → s0(quot(minus(x)))
le → le
rand0(x) → rand0(s0(x))
if_minus(s0(x)) → 00
minus(00) → 00
quot(00) → 00
if_minus(s0(x)) → s0(minus(x))
log0(s0(s0(x))) → s0(log0(s0(quot(x))))

(24) Obligation:

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

IF_MINUS(s0(x), y) → MINUS0(x, y)

The TRS R consists of the following rules:

minus(s0(x)) → if_minus(s0(x))
quot(s0(x)) → s0(quot(minus(x)))
le → le
rand0(x) → rand0(s0(x))
if_minus(s0(x)) → 00
minus(00) → 00
quot(00) → 00
if_minus(s0(x)) → s0(minus(x))
log0(s0(s0(x))) → s0(log0(s0(quot(x))))

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

(25) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(26) TRUE

(27) Obligation:

Relative ADP Problem with
absolute ADPs:

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

and relative ADPs:

minus(s(x), y) → if_minus(le(s(x), y), s(x), y)
log(s(0)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
le(s(x), s(y)) → le(x, y)
le(s(x), 0) → false
rand(x) → rand(s(x))
if_minus(true, s(x), y) → 0
minus(0, y) → 0
quot(0, s(y)) → 0
if_minus(false, s(x), y) → s(minus(x, y))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
le(0, y) → true
rand(x) → x

(28) 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 = true = QUOT_2 = log_1 = le_2 = 0, 1 0 = minus_2 = 1 if_minus_3 = 0, 2 rand_1 = quot_2 = 1 false =
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)  =  x1
if_minus(x1, x2, x3)  =  x2
le(x1, x2)  =  le
0  =  0
false  =  false
true  =  true
quot(x1, x2)  =  x1

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

QUOT2 > [s1, le, true] > false > 0

Status:
QUOT2: [1,2]
s1: multiset
le: multiset
0: multiset
false: multiset
true: multiset

(29) 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:

minus(s0(x)) → if_minus(s0(x))
log0(s0(00)) → 00
quot(s0(x)) → s0(quot(minus(x)))
le → le
le → false0
rand0(x) → rand0(s0(x))
if_minus(s0(x)) → 00
minus(00) → 00
quot(00) → 00
if_minus(s0(x)) → s0(minus(x))
log0(s0(s0(x))) → s0(log0(s0(quot(x))))
le → true0
rand0(x) → x

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

(30) 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:

le → true0
rand0(x) → x

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0   
POL(QUOT0(x1, x2)) = x1 + x2   
POL(false0) = 2   
POL(if_minus(x1)) = x1   
POL(le) = 2   
POL(log0(x1)) = 2·x1   
POL(minus(x1)) = x1   
POL(quot(x1)) = x1   
POL(rand0(x1)) = 2 + x1   
POL(s0(x1)) = x1   
POL(true0) = 0   

(31) 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:

minus(s0(x)) → if_minus(s0(x))
log0(s0(00)) → 00
quot(s0(x)) → s0(quot(minus(x)))
le → le
le → false0
rand0(x) → rand0(s0(x))
if_minus(s0(x)) → 00
minus(00) → 00
quot(00) → 00
if_minus(s0(x)) → s0(minus(x))
log0(s0(s0(x))) → s0(log0(s0(quot(x))))

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

(32) 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:

le → false0

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0   
POL(QUOT0(x1, x2)) = x1 + x2   
POL(false0) = 1   
POL(if_minus(x1)) = x1   
POL(le) = 2   
POL(log0(x1)) = x1   
POL(minus(x1)) = x1   
POL(quot(x1)) = x1   
POL(rand0(x1)) = x1   
POL(s0(x1)) = x1   

(33) 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:

minus(s0(x)) → if_minus(s0(x))
log0(s0(00)) → 00
quot(s0(x)) → s0(quot(minus(x)))
le → le
rand0(x) → rand0(s0(x))
if_minus(s0(x)) → 00
minus(00) → 00
quot(00) → 00
if_minus(s0(x)) → s0(minus(x))
log0(s0(s0(x))) → s0(log0(s0(quot(x))))

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

(34) 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)) = x1 + x2   
POL(if_minus(x1)) = x1   
POL(le) = 0   
POL(log0(x1)) = 2 + x1   
POL(minus(x1)) = x1   
POL(quot(x1)) = x1   
POL(rand0(x1)) = x1   
POL(s0(x1)) = x1   

(35) 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:

minus(s0(x)) → if_minus(s0(x))
quot(s0(x)) → s0(quot(minus(x)))
le → le
rand0(x) → rand0(s0(x))
if_minus(s0(x)) → 00
minus(00) → 00
quot(00) → 00
if_minus(s0(x)) → s0(minus(x))
log0(s0(s0(x))) → s0(log0(s0(quot(x))))

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

(36) 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
if_minus(x1)  =  x1
quot(x1)  =  x1
le  =  le
rand0(x1)  =  rand0
00  =  00
log0(x1)  =  log0(x1)

Recursive path order with status [RPO].
Quasi-Precedence:
[s01, rand0, log01] > [QUOT02, 00]
le > [QUOT02, 00]

Status:
QUOT02: [1,2]
s01: multiset
le: multiset
rand0: multiset
00: multiset
log01: multiset


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

minus(s0(x)) → if_minus(s0(x))
quot(s0(x)) → s0(quot(minus(x)))
le → le
rand0(x) → rand0(s0(x))
if_minus(s0(x)) → 00
minus(00) → 00
quot(00) → 00
if_minus(s0(x)) → s0(minus(x))
log0(s0(s0(x))) → s0(log0(s0(quot(x))))

(37) Obligation:

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

minus(s0(x)) → if_minus(s0(x))
quot(s0(x)) → s0(quot(minus(x)))
le → le
rand0(x) → rand0(s0(x))
if_minus(s0(x)) → 00
minus(00) → 00
quot(00) → 00
if_minus(s0(x)) → s0(minus(x))
log0(s0(s0(x))) → s0(log0(s0(quot(x))))

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

(38) PisEmptyProof (EQUIVALENT transformation)

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

(39) YES

(40) Obligation:

Relative ADP Problem with
absolute ADPs:

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

and relative ADPs:

minus(s(x), y) → if_minus(le(s(x), y), s(x), y)
log(s(0)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
le(s(x), s(y)) → le(x, y)
le(s(x), 0) → false
rand(x) → rand(s(x))
if_minus(true, s(x), y) → 0
minus(0, y) → 0
quot(0, s(y)) → 0
if_minus(false, s(x), y) → s(minus(x, y))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
le(0, y) → true
rand(x) → x

(41) 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 = true = log_1 = 0 = le_2 = 0, 1 minus_2 = 1 if_minus_3 = 0, 2 rand_1 = LOG_1 = quot_2 = 1 false =
Found this filtering by looking at the following order that orders at least one DP strictly:Combined order from the following AFS and order.
LOG(x1)  =  x1
s(x1)  =  s(x1)
quot(x1, x2)  =  x1
0  =  0
minus(x1, x2)  =  x1
if_minus(x1, x2, x3)  =  x2
le(x1, x2)  =  le
false  =  false
true  =  true

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

[le, false, true] > [s1, 0]

Status:
s1: multiset
0: multiset
le: multiset
false: multiset
true: multiset

(42) Obligation:

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

LOG0(s0(s0(x))) → LOG0(s0(quot(x)))

The TRS R consists of the following rules:

minus(s0(x)) → if_minus(s0(x))
log0(s0(00)) → 00
quot(s0(x)) → s0(quot(minus(x)))
le → le
le → false0
rand0(x) → rand0(s0(x))
if_minus(s0(x)) → 00
minus(00) → 00
quot(00) → 00
if_minus(s0(x)) → s0(minus(x))
log0(s0(s0(x))) → s0(log0(s0(quot(x))))
le → true0
rand0(x) → x

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

(43) 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:

le → true0
rand0(x) → x

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0   
POL(LOG0(x1)) = 2·x1   
POL(false0) = 2   
POL(if_minus(x1)) = x1   
POL(le) = 2   
POL(log0(x1)) = x1   
POL(minus(x1)) = x1   
POL(quot(x1)) = x1   
POL(rand0(x1)) = 2 + x1   
POL(s0(x1)) = x1   
POL(true0) = 0   

(44) Obligation:

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

LOG0(s0(s0(x))) → LOG0(s0(quot(x)))

The TRS R consists of the following rules:

minus(s0(x)) → if_minus(s0(x))
log0(s0(00)) → 00
quot(s0(x)) → s0(quot(minus(x)))
le → le
le → false0
rand0(x) → rand0(s0(x))
if_minus(s0(x)) → 00
minus(00) → 00
quot(00) → 00
if_minus(s0(x)) → s0(minus(x))
log0(s0(s0(x))) → s0(log0(s0(quot(x))))

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

(45) 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:

le → false0

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0   
POL(LOG0(x1)) = x1   
POL(false0) = 1   
POL(if_minus(x1)) = x1   
POL(le) = 2   
POL(log0(x1)) = 2·x1   
POL(minus(x1)) = x1   
POL(quot(x1)) = x1   
POL(rand0(x1)) = x1   
POL(s0(x1)) = x1   

(46) Obligation:

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

LOG0(s0(s0(x))) → LOG0(s0(quot(x)))

The TRS R consists of the following rules:

minus(s0(x)) → if_minus(s0(x))
log0(s0(00)) → 00
quot(s0(x)) → s0(quot(minus(x)))
le → le
rand0(x) → rand0(s0(x))
if_minus(s0(x)) → 00
minus(00) → 00
quot(00) → 00
if_minus(s0(x)) → s0(minus(x))
log0(s0(s0(x))) → s0(log0(s0(quot(x))))

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

(47) 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(LOG0(x1)) = 2·x1   
POL(if_minus(x1)) = x1   
POL(le) = 0   
POL(log0(x1)) = 1 + x1   
POL(minus(x1)) = x1   
POL(quot(x1)) = x1   
POL(rand0(x1)) = x1   
POL(s0(x1)) = x1   

(48) Obligation:

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

LOG0(s0(s0(x))) → LOG0(s0(quot(x)))

The TRS R consists of the following rules:

minus(s0(x)) → if_minus(s0(x))
quot(s0(x)) → s0(quot(minus(x)))
le → le
rand0(x) → rand0(s0(x))
if_minus(s0(x)) → 00
minus(00) → 00
quot(00) → 00
if_minus(s0(x)) → s0(minus(x))
log0(s0(s0(x))) → s0(log0(s0(quot(x))))

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

(49) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


LOG0(s0(s0(x))) → LOG0(s0(quot(x)))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
POL( minus(x1) ) = x1

POL( s0(x1) ) = 2x1 + 1

POL( if_minus(x1) ) = x1

POL( quot(x1) ) = x1

POL( le ) = 0

POL( rand0(x1) ) = 2

POL( 00 ) = 1

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

POL( LOG0(x1) ) = max{0, 2x1 - 1}


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

minus(s0(x)) → if_minus(s0(x))
quot(s0(x)) → s0(quot(minus(x)))
le → le
rand0(x) → rand0(s0(x))
if_minus(s0(x)) → 00
minus(00) → 00
quot(00) → 00
if_minus(s0(x)) → s0(minus(x))
log0(s0(s0(x))) → s0(log0(s0(quot(x))))

(50) Obligation:

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

minus(s0(x)) → if_minus(s0(x))
quot(s0(x)) → s0(quot(minus(x)))
le → le
rand0(x) → rand0(s0(x))
if_minus(s0(x)) → 00
minus(00) → 00
quot(00) → 00
if_minus(s0(x)) → s0(minus(x))
log0(s0(s0(x))) → s0(log0(s0(quot(x))))

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

(51) PisEmptyProof (EQUIVALENT transformation)

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

(52) YES