YES Termination proof of AProVE_24_mset_a2.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 (⇒, 38 ms)
        ↳7 QDP
        ↳8 MRRProof (⇔, 9 ms)
        ↳9 QDP
        ↳10 PisEmptyProof (⇔, 0 ms)
        ↳11 YES
    ↳12 RelADPP
        ↳13 RelADPCleverAfsProof (⇒, 38 ms)
        ↳14 QDP
        ↳15 MRRProof (⇔, 7 ms)
        ↳16 QDP
        ↳17 MRRProof (⇔, 0 ms)
        ↳18 QDP
        ↳19 PisEmptyProof (⇔, 0 ms)
        ↳20 YES
    ↳21 RelADPP
        ↳22 RelADPReductionPairProof (⇔, 36 ms)
        ↳23 RelADPP
        ↳24 DAbsisEmptyProof (⇔, 0 ms)
        ↳25 YES
    ↳26 RelADPP
        ↳27 RelADPReductionPairProof (⇔, 39 ms)
        ↳28 RelADPP
        ↳29 DAbsisEmptyProof (⇔, 0 ms)
        ↳30 YES
    ↳31 RelADPP
        ↳32 RelADPReductionPairProof (⇔, 41 ms)
        ↳33 RelADPP
        ↳34 DAbsisEmptyProof (⇔, 0 ms)
        ↳35 YES
    ↳36 RelADPP
        ↳37 RelADPReductionPairProof (⇔, 39 ms)
        ↳38 RelADPP
        ↳39 DAbsisEmptyProof (⇔, 0 ms)
        ↳40 YES

(0) Obligation:

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

minus(x, o) → x
minus(s(x), s(y)) → minus(x, y)
div(0, s(y)) → 0
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
divL(x, nil) → x
divL(x, cons(y, xs)) → divL(div(x, y), xs)

The relative TRS consists of the following S rules:

consSwap(x, xs) → cons(x, xs)
divL(z, cons(x, xs)) → divL(z, consSwap(x, xs))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))

(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, o) → x
minus(s(x), s(y)) → MINUS(x, y)
div(0, s(y)) → 0
div(s(x), s(y)) → s(DIV(minus(x, y), s(y)))
div(s(x), s(y)) → s(div(MINUS(x, y), s(y)))
divL(x, nil) → x
divL(x, cons(y, xs)) → DIVL(div(x, y), xs)
divL(x, cons(y, xs)) → divL(DIV(x, y), xs)

and relative ADPs:

consSwap(x, xs) → cons(x, xs)
divL(z, cons(x, xs)) → DIVL(z, CONSSWAP(x, xs))
consSwap(x, cons(y, xs)) → cons(y, CONSSWAP(x, xs))

(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,
3 Lassos,
Result: This relative DT problem is equivalent to 6 subproblems.

(4) Complex Obligation (AND)

(5) Obligation:

Relative ADP Problem with
absolute ADPs:

minus(s(x), s(y)) → MINUS(x, y)

and relative ADPs:

div(0, s(y)) → 0
consSwap(x, xs) → cons(x, xs)
divL(x, cons(y, xs)) → divL(div(x, y), xs)
minus(x, o) → x
divL(z, cons(x, xs)) → divL(z, consSwap(x, xs))
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
divL(x, nil) → 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 = 1 o = div_2 = divL_2 = 0 = minus_2 = 1 cons_2 = consSwap_2 = nil =
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(x1)
s(x1)  =  s(x1)
div(x1, x2)  =  div(x1, x2)
minus(x1, x2)  =  minus(x1)
0  =  0
o  =  o

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

div2 > s1 > minus1
div2 > 0

Status:
MINUS1: multiset
s1: multiset
div2: [2,1]
minus1: multiset
0: multiset
o: multiset

(7) Obligation:

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

MINUS(s0(x)) → MINUS(x)

The TRS R consists of the following rules:

div0(00, s0(y)) → 00
consSwap0(x, xs) → cons0(x, xs)
divL0(x, cons0(y, xs)) → divL0(div0(x, y), xs)
minus(x) → x
divL0(z, cons0(x, xs)) → divL0(z, consSwap0(x, xs))
div0(s0(x), s0(y)) → s0(div0(minus(x), s0(y)))
minus(s0(x)) → minus(x)
consSwap0(x, cons0(y, xs)) → cons0(y, consSwap0(x, xs))
divL0(x, nil0) → 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 dependency pairs:

MINUS(s0(x)) → MINUS(x)

Strictly oriented rules of the TRS R:

div0(00, s0(y)) → 00
divL0(x, cons0(y, xs)) → divL0(div0(x, y), xs)
minus(s0(x)) → minus(x)
divL0(x, nil0) → x

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0   
POL(MINUS(x1)) = x1   
POL(cons0(x1, x2)) = 2 + x1 + x2   
POL(consSwap0(x1, x2)) = 2 + x1 + x2   
POL(div0(x1, x2)) = x1 + 2·x2   
POL(divL0(x1, x2)) = 2 + x1 + 2·x2   
POL(minus(x1)) = x1   
POL(nil0) = 0   
POL(s0(x1)) = 1 + x1   

(9) Obligation:

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

consSwap0(x, xs) → cons0(x, xs)
minus(x) → x
divL0(z, cons0(x, xs)) → divL0(z, consSwap0(x, xs))
div0(s0(x), s0(y)) → s0(div0(minus(x), s0(y)))
consSwap0(x, cons0(y, xs)) → cons0(y, consSwap0(x, xs))

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

(10) PisEmptyProof (EQUIVALENT transformation)

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

(11) YES

(12) Obligation:

Relative ADP Problem with
absolute ADPs:

div(s(x), s(y)) → s(DIV(minus(x, y), s(y)))

and relative ADPs:

div(0, s(y)) → 0
consSwap(x, xs) → cons(x, xs)
divL(x, cons(y, xs)) → divL(div(x, y), xs)
minus(x, o) → x
divL(z, cons(x, xs)) → divL(z, consSwap(x, xs))
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
minus(s(x), s(y)) → minus(x, y)
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
divL(x, nil) → x

(13) 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 = o = div_2 = 1 divL_2 = 0 = minus_2 = 1 cons_2 = consSwap_2 = DIV_2 = nil =
Found this filtering by looking at the following order that orders at least one DP strictly:Combined order from the following AFS and order.
DIV(x1, x2)  =  DIV(x1, x2)
s(x1)  =  s(x1)
minus(x1, x2)  =  x1
o  =  o
div(x1, x2)  =  div(x1)
0  =  0

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

[s1, div1, 0] > DIV2

Status:
DIV2: multiset
s1: [1]
o: multiset
div1: [1]
0: multiset

(14) Obligation:

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

DIV0(s0(x), s0(y)) → DIV0(minus(x), s0(y))

The TRS R consists of the following rules:

div(00) → 00
consSwap0(x, xs) → cons0(x, xs)
divL0(x, cons0(y, xs)) → divL0(div(x), xs)
minus(x) → x
divL0(z, cons0(x, xs)) → divL0(z, consSwap0(x, xs))
div(s0(x)) → s0(div(minus(x)))
minus(s0(x)) → minus(x)
consSwap0(x, cons0(y, xs)) → cons0(y, consSwap0(x, xs))
divL0(x, nil0) → x

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

(15) 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:

divL0(x, nil0) → x

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0   
POL(DIV0(x1, x2)) = x1 + x2   
POL(cons0(x1, x2)) = x1 + x2   
POL(consSwap0(x1, x2)) = x1 + x2   
POL(div(x1)) = x1   
POL(divL0(x1, x2)) = 2 + x1 + 2·x2   
POL(minus(x1)) = x1   
POL(nil0) = 0   
POL(s0(x1)) = x1   

(16) Obligation:

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

DIV0(s0(x), s0(y)) → DIV0(minus(x), s0(y))

The TRS R consists of the following rules:

div(00) → 00
consSwap0(x, xs) → cons0(x, xs)
divL0(x, cons0(y, xs)) → divL0(div(x), xs)
minus(x) → x
divL0(z, cons0(x, xs)) → divL0(z, consSwap0(x, xs))
div(s0(x)) → s0(div(minus(x)))
minus(s0(x)) → minus(x)
consSwap0(x, cons0(y, xs)) → cons0(y, consSwap0(x, xs))

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 dependency pairs:

DIV0(s0(x), s0(y)) → DIV0(minus(x), s0(y))

Strictly oriented rules of the TRS R:

div(00) → 00
minus(s0(x)) → minus(x)

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 2   
POL(DIV0(x1, x2)) = 2·x1 + x2   
POL(cons0(x1, x2)) = 1 + x1 + x2   
POL(consSwap0(x1, x2)) = 1 + x1 + x2   
POL(div(x1)) = 1 + x1   
POL(divL0(x1, x2)) = 2·x1 + 2·x2   
POL(minus(x1)) = x1   
POL(s0(x1)) = 1 + x1   

(18) Obligation:

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

consSwap0(x, xs) → cons0(x, xs)
divL0(x, cons0(y, xs)) → divL0(div(x), xs)
minus(x) → x
divL0(z, cons0(x, xs)) → divL0(z, consSwap0(x, xs))
div(s0(x)) → s0(div(minus(x)))
consSwap0(x, cons0(y, xs)) → cons0(y, consSwap0(x, xs))

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

(19) PisEmptyProof (EQUIVALENT transformation)

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

(20) YES

(21) Obligation:

Relative ADP Problem with
absolute ADPs:

divL(x, cons(y, xs)) → DIVL(div(x, y), xs)

and relative ADPs:

div(0, s(y)) → 0
consSwap(x, xs) → cons(x, xs)
divL(x, cons(y, xs)) → divL(div(x, y), xs)
minus(x, o) → x
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
minus(s(x), s(y)) → minus(x, y)
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
divL(z, cons(x, xs)) → DIVL(z, CONSSWAP(x, xs))
divL(x, nil) → x

(22) 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:


divL(x, cons(y, xs)) → DIVL(div(x, y), xs)

Relative ADPs:

div(0, s(y)) → 0
consSwap(x, xs) → cons(x, xs)
divL(x, cons(y, xs)) → divL(div(x, y), xs)
minus(x, o) → x
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
minus(s(x), s(y)) → minus(x, y)
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
divL(x, nil) → x


The remaining rules can at least be oriented weakly:

Ordered with Polynomial interpretation [POLO]:

POL(0) = 0   
POL(CONSSWAP(x1, x2)) = 0   
POL(DIV(x1, x2)) = 2 + 2·x1·x2   
POL(DIVL(x1, x2)) = 2 + x1 + 3·x2   
POL(MINUS(x1, x2)) = 0   
POL(cons(x1, x2)) = 3 + 3·x1 + x2   
POL(consSwap(x1, x2)) = 3 + 3·x1 + x2   
POL(div(x1, x2)) = x1   
POL(divL(x1, x2)) = 2·x1   
POL(minus(x1, x2)) = x1   
POL(nil) = 0   
POL(o) = 2   
POL(s(x1)) = 1 + x1   

(23) Obligation:

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

div(0, s(y)) → 0
consSwap(x, xs) → cons(x, xs)
divL(x, cons(y, xs)) → divL(div(x, y), xs)
minus(x, o) → x
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
minus(s(x), s(y)) → minus(x, y)
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
divL(z, cons(x, xs)) → DIVL(z, CONSSWAP(x, xs))
divL(x, nil) → x

(24) DAbsisEmptyProof (EQUIVALENT transformation)

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

(25) YES

(26) Obligation:

Relative ADP Problem with
absolute ADPs:

divL(x, cons(y, xs)) → DIVL(div(x, y), xs)

and relative ADPs:

div(0, s(y)) → 0
consSwap(x, xs) → cons(x, xs)
divL(x, cons(y, xs)) → divL(div(x, y), xs)
minus(x, o) → x
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
minus(s(x), s(y)) → minus(x, y)
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
divL(z, cons(x, xs)) → DIVL(z, CONSSWAP(x, xs))
divL(x, nil) → x

(27) 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:


divL(x, cons(y, xs)) → DIVL(div(x, y), xs)

Relative ADPs:

div(0, s(y)) → 0
consSwap(x, xs) → cons(x, xs)
divL(x, cons(y, xs)) → divL(div(x, y), xs)
minus(x, o) → x
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
minus(s(x), s(y)) → minus(x, y)
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
divL(x, nil) → x


The remaining rules can at least be oriented weakly:

Ordered with Polynomial interpretation [POLO]:

POL(0) = 0   
POL(CONSSWAP(x1, x2)) = 0   
POL(DIV(x1, x2)) = 2 + 2·x1·x2   
POL(DIVL(x1, x2)) = 2 + x1 + 3·x2   
POL(MINUS(x1, x2)) = 0   
POL(cons(x1, x2)) = 3 + 3·x1 + x2   
POL(consSwap(x1, x2)) = 3 + 3·x1 + x2   
POL(div(x1, x2)) = x1   
POL(divL(x1, x2)) = 2·x1   
POL(minus(x1, x2)) = x1   
POL(nil) = 0   
POL(o) = 2   
POL(s(x1)) = 1 + x1   

(28) Obligation:

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

div(0, s(y)) → 0
consSwap(x, xs) → cons(x, xs)
divL(x, cons(y, xs)) → divL(div(x, y), xs)
minus(x, o) → x
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
minus(s(x), s(y)) → minus(x, y)
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
divL(z, cons(x, xs)) → DIVL(z, CONSSWAP(x, xs))
divL(x, nil) → x

(29) DAbsisEmptyProof (EQUIVALENT transformation)

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

(30) YES

(31) Obligation:

Relative ADP Problem with
absolute ADPs:

divL(x, cons(y, xs)) → divL(DIV(x, y), xs)

and relative ADPs:

div(0, s(y)) → 0
consSwap(x, xs) → cons(x, xs)
divL(x, cons(y, xs)) → divL(div(x, y), xs)
minus(x, o) → x
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
minus(s(x), s(y)) → minus(x, y)
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
divL(z, cons(x, xs)) → DIVL(z, CONSSWAP(x, xs))
divL(x, nil) → x

(32) 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:


divL(x, cons(y, xs)) → divL(DIV(x, y), xs)

Relative ADPs:

div(0, s(y)) → 0
consSwap(x, xs) → cons(x, xs)
divL(x, cons(y, xs)) → divL(div(x, y), xs)
minus(x, o) → x
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
minus(s(x), s(y)) → minus(x, y)
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
divL(x, nil) → x


The remaining rules can at least be oriented weakly:

Ordered with Polynomial interpretation [POLO]:

POL(0) = 0   
POL(CONSSWAP(x1, x2)) = 0   
POL(DIV(x1, x2)) = 1   
POL(DIVL(x1, x2)) = 2 + 2·x1 + x2   
POL(MINUS(x1, x2)) = 2·x1·x2   
POL(cons(x1, x2)) = 2 + 3·x2   
POL(consSwap(x1, x2)) = 2 + 3·x2   
POL(div(x1, x2)) = 2 + x1   
POL(divL(x1, x2)) = 2 + 2·x1 + 2·x2   
POL(minus(x1, x2)) = x1   
POL(nil) = 0   
POL(o) = 0   
POL(s(x1)) = 3 + x1   

(33) Obligation:

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

div(0, s(y)) → 0
consSwap(x, xs) → cons(x, xs)
divL(x, cons(y, xs)) → divL(div(x, y), xs)
minus(x, o) → x
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
minus(s(x), s(y)) → minus(x, y)
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
divL(z, cons(x, xs)) → DIVL(z, CONSSWAP(x, xs))
divL(x, nil) → x

(34) DAbsisEmptyProof (EQUIVALENT transformation)

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

(35) YES

(36) Obligation:

Relative ADP Problem with
absolute ADPs:

divL(x, nil) → x

and relative ADPs:

div(0, s(y)) → 0
consSwap(x, xs) → cons(x, xs)
divL(x, cons(y, xs)) → divL(div(x, y), xs)
minus(x, o) → x
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
minus(s(x), s(y)) → minus(x, y)
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
divL(z, cons(x, xs)) → DIVL(z, CONSSWAP(x, xs))

(37) 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:


divL(x, nil) → x

Relative ADPs:

div(0, s(y)) → 0
consSwap(x, xs) → cons(x, xs)
divL(x, cons(y, xs)) → divL(div(x, y), xs)
minus(x, o) → x
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
minus(s(x), s(y)) → minus(x, y)
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))


The remaining rules can at least be oriented weakly:

Ordered with Polynomial interpretation [POLO]:

POL(0) = 0   
POL(CONSSWAP(x1, x2)) = 0   
POL(DIV(x1, x2)) = 0   
POL(DIVL(x1, x2)) = 1 + 3·x1 + x2   
POL(MINUS(x1, x2)) = 2   
POL(cons(x1, x2)) = 2 + 3·x2   
POL(consSwap(x1, x2)) = 2 + 3·x2   
POL(div(x1, x2)) = x1   
POL(divL(x1, x2)) = 2 + 2·x1   
POL(minus(x1, x2)) = x1   
POL(nil) = 0   
POL(o) = 0   
POL(s(x1)) = 2 + 3·x1   

(38) Obligation:

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

div(0, s(y)) → 0
consSwap(x, xs) → cons(x, xs)
divL(x, cons(y, xs)) → divL(div(x, y), xs)
minus(x, o) → x
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
minus(s(x), s(y)) → minus(x, y)
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
divL(z, cons(x, xs)) → DIVL(z, CONSSWAP(x, xs))
divL(x, nil) → x

(39) DAbsisEmptyProof (EQUIVALENT transformation)

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

(40) YES