YES Termination proof of INVY_15_3.8a_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 (⇒, 51 ms)
        ↳7 QDP
        ↳8 MRRProof (⇔, 20 ms)
        ↳9 QDP
        ↳10 MRRProof (⇔, 0 ms)
        ↳11 QDP
        ↳12 QDPOrderProof (⇔, 12 ms)
        ↳13 QDP
        ↳14 PisEmptyProof (⇔, 0 ms)
        ↳15 YES
    ↳16 RelADPP
        ↳17 RelADPCleverAfsProof (⇒, 41 ms)
        ↳18 QDP
        ↳19 MRRProof (⇔, 20 ms)
        ↳20 QDP
        ↳21 QDPOrderProof (⇔, 0 ms)
        ↳22 QDP
        ↳23 PisEmptyProof (⇔, 0 ms)
        ↳24 YES
    ↳25 RelADPP
        ↳26 RelADPCleverAfsProof (⇒, 35 ms)
        ↳27 QDP
        ↳28 MRRProof (⇔, 18 ms)
        ↳29 QDP
        ↳30 QDPOrderProof (⇔, 0 ms)
        ↳31 QDP
        ↳32 PisEmptyProof (⇔, 0 ms)
        ↳33 YES

(0) Obligation:

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

pred(s(x)) → x
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)))
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:

pred(s(x)) → x
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)))
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:
3 SCCs with nodes from P_abs,
0 Lassos,
Result: This relative DT problem is equivalent to 3 subproblems.

(4) Complex Obligation (AND)

(5) Obligation:

Relative ADP Problem with
absolute ADPs:

minus(x, s(y)) → pred(MINUS(x, y))

and relative ADPs:

log(s(0)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(x, s(y)) → pred(minus(x, y))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
rand(x) → rand(s(x))
rand(x) → x
pred(s(x)) → x
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 = MINUS_2 = 0 log_1 = pred_1 = 0 = minus_2 = 1 rand_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.
MINUS(x1, x2)  =  x2
s(x1)  =  s(x1)
quot(x1, x2)  =  quot(x1, x2)
minus(x1, x2)  =  minus(x1)
pred(x1)  =  x1
0  =  0

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

quot2 > s1 > [minus1, 0]

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

log0(s0(00)) → 00
quot0(s0(x), s0(y)) → s0(quot0(minus(x), s0(y)))
minus(x) → pred0(minus(x))
log0(s0(s0(x))) → s0(log0(s0(quot0(x, s0(s0(00))))))
minus(x) → x
rand0(x) → rand0(s0(x))
rand0(x) → x
pred0(s0(x)) → x
quot0(00, s0(y)) → 00

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:

rand0(x) → x

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0   
POL(MINUS(x1)) = x1   
POL(log0(x1)) = 2·x1   
POL(minus(x1)) = x1   
POL(pred0(x1)) = x1   
POL(quot0(x1, x2)) = x1 + 2·x2   
POL(rand0(x1)) = 2 + x1   
POL(s0(x1)) = 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:

log0(s0(00)) → 00
quot0(s0(x), s0(y)) → s0(quot0(minus(x), s0(y)))
minus(x) → pred0(minus(x))
log0(s0(s0(x))) → s0(log0(s0(quot0(x, s0(s0(00))))))
minus(x) → x
rand0(x) → rand0(s0(x))
pred0(s0(x)) → x
quot0(00, s0(y)) → 00

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:

log0(s0(00)) → 00

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0   
POL(MINUS(x1)) = x1   
POL(log0(x1)) = 1 + 2·x1   
POL(minus(x1)) = x1   
POL(pred0(x1)) = x1   
POL(quot0(x1, x2)) = x1 + x2   
POL(rand0(x1)) = x1   
POL(s0(x1)) = x1   

(11) 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:

quot0(s0(x), s0(y)) → s0(quot0(minus(x), s0(y)))
minus(x) → pred0(minus(x))
log0(s0(s0(x))) → s0(log0(s0(quot0(x, s0(s0(00))))))
minus(x) → x
rand0(x) → rand0(s0(x))
pred0(s0(x)) → x
quot0(00, s0(y)) → 00

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.


MINUS(s0(y)) → MINUS(y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MINUS(x1)  =  x1
s0(x1)  =  s0(x1)
quot0(x1, x2)  =  x1
minus(x1)  =  x1
pred0(x1)  =  x1
log0(x1)  =  log0(x1)
00  =  00
rand0(x1)  =  rand0

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

Status:
s01: multiset
log01: multiset
00: multiset
rand0: []


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

quot0(s0(x), s0(y)) → s0(quot0(minus(x), s0(y)))
minus(x) → pred0(minus(x))
log0(s0(s0(x))) → s0(log0(s0(quot0(x, s0(s0(00))))))
minus(x) → x
rand0(x) → rand0(s0(x))
pred0(s0(x)) → x
quot0(00, s0(y)) → 00

(13) Obligation:

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

quot0(s0(x), s0(y)) → s0(quot0(minus(x), s0(y)))
minus(x) → pred0(minus(x))
log0(s0(s0(x))) → s0(log0(s0(quot0(x, s0(s0(00))))))
minus(x) → x
rand0(x) → rand0(s0(x))
pred0(s0(x)) → x
quot0(00, s0(y)) → 00

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:

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

and relative ADPs:

log(s(0)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(x, s(y)) → pred(minus(x, y))
log(s(s(x))) → s(log(s(quot(x, s(s(0))))))
minus(x, 0) → x
rand(x) → rand(s(x))
rand(x) → x
pred(s(x)) → x
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 = QUOT_2 = 1 log_1 = 0 = pred_1 = minus_2 = 1 rand_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.
QUOT(x1, x2)  =  QUOT(x1)
s(x1)  =  s(x1)
minus(x1, x2)  =  x1
pred(x1)  =  x1
0  =  0
quot(x1, x2)  =  quot(x1)

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

[QUOT1, s1, 0, quot1]

Status:
QUOT1: multiset
s1: multiset
0: multiset
quot1: multiset

(18) Obligation:

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

QUOT(s0(x)) → QUOT(minus(x))

The TRS R consists of the following rules:

log0(s0(00)) → 00
quot(s0(x)) → s0(quot(minus(x)))
minus(x) → pred0(minus(x))
log0(s0(s0(x))) → s0(log0(s0(quot(x))))
minus(x) → x
rand0(x) → rand0(s0(x))
rand0(x) → x
pred0(s0(x)) → x
quot(00) → 00

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
rand0(x) → x

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 1   
POL(QUOT(x1)) = x1   
POL(log0(x1)) = 2 + x1   
POL(minus(x1)) = x1   
POL(pred0(x1)) = x1   
POL(quot(x1)) = x1   
POL(rand0(x1)) = 2 + x1   
POL(s0(x1)) = x1   

(20) Obligation:

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

QUOT(s0(x)) → QUOT(minus(x))

The TRS R consists of the following rules:

quot(s0(x)) → s0(quot(minus(x)))
minus(x) → pred0(minus(x))
log0(s0(s0(x))) → s0(log0(s0(quot(x))))
minus(x) → x
rand0(x) → rand0(s0(x))
pred0(s0(x)) → x
quot(00) → 00

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

(21) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


QUOT(s0(x)) → QUOT(minus(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
QUOT(x1)  =  QUOT(x1)
s0(x1)  =  s0(x1)
minus(x1)  =  x1
quot(x1)  =  x1
pred0(x1)  =  x1
log0(x1)  =  x1
rand0(x1)  =  rand0
00  =  00

Recursive path order with status [RPO].
Quasi-Precedence:
[QUOT1, s01, rand0]

Status:
QUOT1: multiset
s01: multiset
rand0: multiset
00: multiset


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

quot(s0(x)) → s0(quot(minus(x)))
minus(x) → pred0(minus(x))
log0(s0(s0(x))) → s0(log0(s0(quot(x))))
minus(x) → x
rand0(x) → rand0(s0(x))
pred0(s0(x)) → x
quot(00) → 00

(22) Obligation:

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

quot(s0(x)) → s0(quot(minus(x)))
minus(x) → pred0(minus(x))
log0(s0(s0(x))) → s0(log0(s0(quot(x))))
minus(x) → x
rand0(x) → rand0(s0(x))
pred0(s0(x)) → x
quot(00) → 00

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

(23) PisEmptyProof (EQUIVALENT transformation)

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

(24) YES

(25) Obligation:

Relative ADP Problem with
absolute ADPs:

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

and relative ADPs:

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

(26) 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 = log_1 = 0 = pred_1 = minus_2 = 1 rand_1 = LOG_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.
LOG(x1)  =  LOG(x1)
s(x1)  =  s(x1)
quot(x1, x2)  =  x1
0  =  0
minus(x1, x2)  =  x1
pred(x1)  =  x1

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

[LOG1, s1, 0]

Status:
LOG1: multiset
s1: multiset
0: multiset

(27) 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:

log0(s0(00)) → 00
quot(s0(x)) → s0(quot(minus(x)))
minus(x) → pred0(minus(x))
log0(s0(s0(x))) → s0(log0(s0(quot(x))))
minus(x) → x
rand0(x) → rand0(s0(x))
rand0(x) → x
pred0(s0(x)) → x
quot(00) → 00

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

(28) 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
rand0(x) → x

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 1   
POL(LOG0(x1)) = x1   
POL(log0(x1)) = 2 + x1   
POL(minus(x1)) = x1   
POL(pred0(x1)) = x1   
POL(quot(x1)) = x1   
POL(rand0(x1)) = 2 + x1   
POL(s0(x1)) = x1   

(29) 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:

quot(s0(x)) → s0(quot(minus(x)))
minus(x) → pred0(minus(x))
log0(s0(s0(x))) → s0(log0(s0(quot(x))))
minus(x) → x
rand0(x) → rand0(s0(x))
pred0(s0(x)) → x
quot(00) → 00

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

(30) 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: Combined order from the following AFS and order.
LOG0(x1)  =  LOG0(x1)
s0(x1)  =  s0(x1)
quot(x1)  =  x1
minus(x1)  =  x1
pred0(x1)  =  x1
log0(x1)  =  log0(x1)
rand0(x1)  =  rand0
00  =  00

Recursive path order with status [RPO].
Quasi-Precedence:
log01 > [LOG01, s01, rand0]

Status:
LOG01: multiset
s01: multiset
log01: multiset
rand0: []
00: multiset


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

quot(s0(x)) → s0(quot(minus(x)))
minus(x) → pred0(minus(x))
log0(s0(s0(x))) → s0(log0(s0(quot(x))))
minus(x) → x
rand0(x) → rand0(s0(x))
pred0(s0(x)) → x
quot(00) → 00

(31) Obligation:

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

quot(s0(x)) → s0(quot(minus(x)))
minus(x) → pred0(minus(x))
log0(s0(s0(x))) → s0(log0(s0(quot(x))))
minus(x) → x
rand0(x) → rand0(s0(x))
pred0(s0(x)) → x
quot(00) → 00

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

(32) PisEmptyProof (EQUIVALENT transformation)

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

(33) YES