Termination proof of AProVE_24_mset

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 (⇒, 54 ms)

        ↳7 QDP

        ↳8 MRRProof (⇔, 0 ms)

        ↳9 QDP

        ↳10 MRRProof (⇔, 0 ms)

        ↳11 QDP

        ↳12 PisEmptyProof (⇔, 0 ms)

        ↳13 YES

    ↳14 RelADPP

        ↳15 RelADPCleverAfsProof (⇒, 46 ms)

        ↳16 QDP

        ↳17 MRRProof (⇔, 0 ms)

        ↳18 QDP

        ↳19 MRRProof (⇔, 0 ms)

        ↳20 QDP

        ↳21 PisEmptyProof (⇔, 0 ms)

        ↳22 YES

    ↳23 RelADPP

        ↳24 RelADPReductionPairProof (⇔, 51 ms)

        ↳25 RelADPP

        ↳26 DAbsisEmptyProof (⇔, 0 ms)

        ↳27 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))
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:

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

pred(s(x)) → x
minus(x, 0) → x
minus(x, s(y)) → PRED(minus(x, y))
minus(x, s(y)) → pred(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:

divL(z, cons(x, cons(y, xs))) → DIVL(z, cons(y, cons(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,
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:

div(0, s(y)) → 0
divL(x, cons(y, xs)) → divL(div(x, y), xs)
divL(z, cons(x, cons(y, xs))) → divL(z, cons(y, cons(x, xs)))
minus(x, s(y)) → pred(minus(x, y))
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
minus(x, 0) → x
pred(s(x)) → x
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 = 0 div_2 = 1 divL_2 = pred_1 = 0 = minus_2 = 1 cons_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(x2)
s(x1) = s(x1)
div(x1, x2) = div(x1)
minus(x1, x2) = x1
0 = 0
pred(x1) = x1

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

[s₁, div₁]

Status:

MINUS₁: multiset
s₁: multiset
div₁: 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:

div(00) → 00
divL0(x, cons0(y, xs)) → divL0(div(x), xs)
divL0(z, cons0(x, cons0(y, xs))) → divL0(z, cons0(y, cons0(x, xs)))
minus(x) → pred0(minus(x))
div(s0(x)) → s0(div(minus(x)))
minus(x) → x
pred0(s0(x)) → x
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 rules of the TRS R:

divL0(x, cons0(y, xs)) → divL0(div(x), xs)
divL0(x, nil0) → x

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0
POL(MINUS(x₁)) = x₁
POL(cons0(x₁, x₂)) = 1 + x₁ + x₂
POL(div(x₁)) = x₁
POL(divL0(x₁, x₂)) = 2 + 2·x₁ + 2·x₂
POL(minus(x₁)) = x₁
POL(nil0) = 0
POL(pred0(x₁)) = 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:

div(00) → 00
divL0(z, cons0(x, cons0(y, xs))) → divL0(z, cons0(y, cons0(x, xs)))
minus(x) → pred0(minus(x))
div(s0(x)) → s0(div(minus(x)))
minus(x) → x
pred0(s0(x)) → x

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

(10) MRRProof (EQUIVALENT transformation)

MINUS(s0(y)) → MINUS(y)

Strictly oriented rules of the TRS R:

div(s0(x)) → s0(div(minus(x)))
pred0(s0(x)) → x

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0
POL(MINUS(x₁)) = x₁
POL(cons0(x₁, x₂)) = x₁ + x₂
POL(div(x₁)) = 2·x₁
POL(divL0(x₁, x₂)) = x₁ + 2·x₂
POL(minus(x₁)) = x₁
POL(pred0(x₁)) = x₁
POL(s0(x₁)) = 1 + x₁

(11) Obligation:

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

div(00) → 00
divL0(z, cons0(x, cons0(y, xs))) → divL0(z, cons0(y, cons0(x, xs)))
minus(x) → pred0(minus(x))
minus(x) → x

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:

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

and relative ADPs:

div(0, s(y)) → 0
divL(x, cons(y, xs)) → divL(div(x, y), xs)
divL(z, cons(x, cons(y, xs))) → divL(z, cons(y, cons(x, xs)))
minus(x, s(y)) → pred(minus(x, y))
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
minus(x, 0) → x
pred(s(x)) → x
divL(x, nil) → 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 = div_2 = 1 divL_2 = 0 = pred_1 = minus_2 = 1 cons_2 = DIV_2 = 1 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)
s(x1) = s(x1)
minus(x1, x2) = x1
pred(x1) = x1
0 = 0
div(x1, x2) = div(x1)

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

[DIV₁, s₁, 0, div₁]

Status:

DIV₁: multiset
s₁: multiset
0: multiset
div₁: multiset

(16) Obligation:

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

DIV(s0(x)) → DIV(minus(x))

The TRS R consists of the following rules:

div(00) → 00
divL0(x, cons0(y, xs)) → divL0(div(x), xs)
divL0(z, cons0(x, cons0(y, xs))) → divL0(z, cons0(y, cons0(x, xs)))
minus(x) → pred0(minus(x))
div(s0(x)) → s0(div(minus(x)))
minus(x) → x
pred0(s0(x)) → x
divL0(x, nil0) → x

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

(17) MRRProof (EQUIVALENT transformation)

divL0(x, cons0(y, xs)) → divL0(div(x), xs)
divL0(x, nil0) → x

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0
POL(DIV(x₁)) = x₁
POL(cons0(x₁, x₂)) = 1 + 2·x₁ + x₂
POL(div(x₁)) = x₁
POL(divL0(x₁, x₂)) = 2 + 2·x₁ + 2·x₂
POL(minus(x₁)) = x₁
POL(nil0) = 0
POL(pred0(x₁)) = x₁
POL(s0(x₁)) = x₁

(18) Obligation:

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

DIV(s0(x)) → DIV(minus(x))

The TRS R consists of the following rules:

div(00) → 00
divL0(z, cons0(x, cons0(y, xs))) → divL0(z, cons0(y, cons0(x, xs)))
minus(x) → pred0(minus(x))
div(s0(x)) → s0(div(minus(x)))
minus(x) → x
pred0(s0(x)) → x

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

(19) MRRProof (EQUIVALENT transformation)

DIV(s0(x)) → DIV(minus(x))

Strictly oriented rules of the TRS R:

minus(x) → x
pred0(s0(x)) → x

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0
POL(DIV(x₁)) = x₁
POL(cons0(x₁, x₂)) = x₁ + x₂
POL(div(x₁)) = 2·x₁
POL(divL0(x₁, x₂)) = x₁ + 2·x₂
POL(minus(x₁)) = 1 + x₁
POL(pred0(x₁)) = x₁
POL(s0(x₁)) = 2 + x₁

(20) Obligation:

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

div(00) → 00
divL0(z, cons0(x, cons0(y, xs))) → divL0(z, cons0(y, cons0(x, xs)))
minus(x) → pred0(minus(x))
div(s0(x)) → s0(div(minus(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:

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

and relative ADPs:

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

(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:
Absolute ADPs:

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

Relative ADPs:

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

The remaining rules can at least be oriented weakly:

Ordered with Polynomial interpretation [POLO]:

POL(0) = 0
POL(DIV(x₁, x₂)) = 2
POL(DIVL(x₁, x₂)) = 3·x₁ + 3·x₂
POL(MINUS(x₁, x₂)) = 2·x₁ + 2·x₁·x₂
POL(PRED(x₁)) = 2 + x₁
POL(cons(x₁, x₂)) = 3 + 2·x₂
POL(div(x₁, x₂)) = 2
POL(divL(x₁, x₂)) = 2 + 2·x₁ + 2·x₂
POL(minus(x₁, x₂)) = 2 + 2·x₁ + 2·x₂
POL(nil) = 0
POL(pred(x₁)) = x₁
POL(s(x₁)) = x₁

(25) Obligation:

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

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

(26) DAbsisEmptyProof (EQUIVALENT transformation)

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

(0) Obligation:

(1) RelTRStoRelADPProof (EQUIVALENT transformation)

(2) Obligation:

(3) RelADPDepGraphProof (EQUIVALENT transformation)

(4) Complex Obligation (AND)

(5) Obligation:

(6) RelADPCleverAfsProof (SOUND transformation)

(7) Obligation:

(8) MRRProof (EQUIVALENT transformation)

(9) Obligation:

(10) MRRProof (EQUIVALENT transformation)

(11) Obligation:

(12) PisEmptyProof (EQUIVALENT transformation)

(13) YES

(14) Obligation:

(15) RelADPCleverAfsProof (SOUND transformation)

(16) Obligation:

(17) MRRProof (EQUIVALENT transformation)

(18) Obligation:

(19) MRRProof (EQUIVALENT transformation)

(20) Obligation:

(21) PisEmptyProof (EQUIVALENT transformation)

(22) YES

(23) Obligation:

(24) RelADPReductionPairProof (EQUIVALENT transformation)

(25) Obligation:

(26) DAbsisEmptyProof (EQUIVALENT transformation)

(27) YES