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 (⇒, 86 ms)

        ↳7 QDP

        ↳8 MRRProof (⇔, 0 ms)

        ↳9 QDP

        ↳10 MRRProof (⇔, 8 ms)

        ↳11 QDP

        ↳12 MRRProof (⇔, 0 ms)

        ↳13 QDP

        ↳14 PisEmptyProof (⇔, 0 ms)

        ↳15 YES

    ↳16 RelADPP

        ↳17 RelADPCleverAfsProof (⇒, 86 ms)

        ↳18 QDP

        ↳19 MRRProof (⇔, 0 ms)

        ↳20 QDP

        ↳21 MRRProof (⇔, 6 ms)

        ↳22 QDP

        ↳23 DependencyGraphProof (⇔, 0 ms)

        ↳24 TRUE

    ↳25 RelADPP

        ↳26 RelADPCleverAfsProof (⇒, 80 ms)

        ↳27 QDP

        ↳28 MRRProof (⇔, 0 ms)

        ↳29 QDP

        ↳30 MRRProof (⇔, 5 ms)

        ↳31 QDP

        ↳32 MRRProof (⇔, 0 ms)

        ↳33 QDP

        ↳34 PisEmptyProof (⇔, 0 ms)

        ↳35 YES

    ↳36 RelADPP

        ↳37 RelADPReductionPairProof (⇔, 110 ms)

        ↳38 RelADPP

        ↳39 DAbsisEmptyProof (⇔, 0 ms)

        ↳40 YES

    ↳41 RelADPP

        ↳42 RelADPReductionPairProof (⇔, 86 ms)

        ↳43 RelADPP

        ↳44 DAbsisEmptyProof (⇔, 0 ms)

        ↳45 YES

    ↳46 RelADPP

        ↳47 RelADPReductionPairProof (⇔, 83 ms)

        ↳48 RelADPP

        ↳49 DAbsisEmptyProof (⇔, 0 ms)

        ↳50 YES

    ↳51 RelADPP

        ↳52 RelADPReductionPairProof (⇔, 87 ms)

        ↳53 RelADPP

        ↳54 DAbsisEmptyProof (⇔, 0 ms)

        ↳55 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) → ifminus(le(s(x), y), s(x), y)
ifminus(true, s(x), y) → 0
ifminus(false, s(x), y) → s(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:

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) → IFMINUS(le(s(x), y), s(x), y)
minus(s(x), y) → ifminus(LE(s(x), y), s(x), y)
ifminus(true, s(x), y) → 0
ifminus(false, s(x), y) → s(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:
4 SCCs with nodes from P_abs,
3 Lassos,
Result: This relative DT problem is equivalent to 7 subproblems.

(4) Complex Obligation (AND)

(5) Obligation:

Relative ADP Problem with
absolute ADPs:

le(s(x), s(y)) → LE(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)
divL(z, cons(x, xs)) → divL(z, consSwap(x, xs))
le(s(x), 0) → false
ifminus(true, s(x), y) → 0
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
minus(0, y) → 0
divL(x, nil) → x
minus(s(x), y) → ifminus(le(s(x), y), s(x), y)
le(0, y) → true
ifminus(false, s(x), y) → s(minus(x, y))

(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:true = ifminus_3 = 0, 2 divL_2 = false = nil = s_1 = div_2 = 1 LE_2 = 1 le_2 = 0, 1 0 = minus_2 = 1 cons_2 = consSwap_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.
LE(x1, x2) = LE(x1)
s(x1) = s(x1)
div(x1, x2) = div(x1)
minus(x1, x2) = x1
ifminus(x1, x2, x3) = x2
le(x1, x2) = le
0 = 0
false = false
true = true

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

[LE₁, s₁, div₁, le, 0, false, true]

Status:

LE₁: multiset
s₁: multiset
div₁: multiset
le: multiset
0: multiset
false: multiset
true: multiset

(7) Obligation:

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

LE(s0(x)) → LE(x)

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)
le → le
divL0(z, cons0(x, xs)) → divL0(z, consSwap0(x, xs))
le → false0
ifminus(s0(x)) → 00
div(s0(x)) → s0(div(minus(x)))
consSwap0(x, cons0(y, xs)) → cons0(y, consSwap0(x, xs))
minus(00) → 00
divL0(x, nil0) → x
minus(s0(x)) → ifminus(s0(x))
le → true0
ifminus(s0(x)) → s0(minus(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:

le → false0
divL0(x, nil0) → x

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0
POL(LE(x₁)) = x₁
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(false0) = 0
POL(ifminus(x₁)) = x₁
POL(le) = 2
POL(minus(x₁)) = x₁
POL(nil0) = 0
POL(s0(x₁)) = 2·x₁
POL(true0) = 2

(9) Obligation:

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

LE(s0(x)) → LE(x)

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)
le → le
divL0(z, cons0(x, xs)) → divL0(z, consSwap0(x, xs))
ifminus(s0(x)) → 00
div(s0(x)) → s0(div(minus(x)))
consSwap0(x, cons0(y, xs)) → cons0(y, consSwap0(x, xs))
minus(00) → 00
minus(s0(x)) → ifminus(s0(x))
le → true0
ifminus(s0(x)) → s0(minus(x))

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

(10) MRRProof (EQUIVALENT transformation)

le → true0

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0
POL(LE(x₁)) = 2·x₁
POL(cons0(x₁, x₂)) = x₁ + x₂
POL(consSwap0(x₁, x₂)) = x₁ + x₂
POL(div(x₁)) = x₁
POL(divL0(x₁, x₂)) = x₁ + 2·x₂
POL(ifminus(x₁)) = x₁
POL(le) = 2
POL(minus(x₁)) = x₁
POL(s0(x₁)) = x₁
POL(true0) = 1

(11) Obligation:

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

LE(s0(x)) → LE(x)

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

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

(12) MRRProof (EQUIVALENT transformation)

LE(s0(x)) → LE(x)

Strictly oriented rules of the TRS R:

div(00) → 00
divL0(x, cons0(y, xs)) → divL0(div(x), xs)
ifminus(s0(x)) → 00

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0
POL(LE(x₁)) = 2·x₁
POL(cons0(x₁, x₂)) = 2 + x₁ + x₂
POL(consSwap0(x₁, x₂)) = 2 + x₁ + x₂
POL(div(x₁)) = 1 + x₁
POL(divL0(x₁, x₂)) = x₁ + 2·x₂
POL(ifminus(x₁)) = x₁
POL(le) = 0
POL(minus(x₁)) = x₁
POL(s0(x₁)) = 1 + x₁

(13) Obligation:

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

consSwap0(x, xs) → cons0(x, xs)
le → le
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))
minus(00) → 00
minus(s0(x)) → ifminus(s0(x))
ifminus(s0(x)) → s0(minus(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:

minus(s(x), y) → IFMINUS(le(s(x), y), s(x), y)
ifminus(false, s(x), y) → s(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)
le(s(x), s(y)) → le(x, y)
divL(z, cons(x, xs)) → divL(z, consSwap(x, xs))
le(s(x), 0) → false
ifminus(true, s(x), y) → 0
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
minus(0, y) → 0
divL(x, nil) → x
minus(s(x), y) → ifminus(le(s(x), y), s(x), y)
le(0, y) → true

(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:MINUS_2 = 1 true = ifminus_3 = 0, 2 divL_2 = IFMINUS_3 = 2 false = nil = s_1 = div_2 = 1 le_2 = 0, 1 0 = minus_2 = 1 cons_2 = consSwap_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) = MINUS(x1)
s(x1) = s(x1)
IFMINUS(x1, x2, x3) = IFMINUS(x1, x2)
le(x1, x2) = le
false = false
0 = 0
true = true
minus(x1, x2) = x1
ifminus(x1, x2, x3) = x2
div(x1, x2) = x1

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

s₁ > [MINUS₁, le, false, true] > [IFMINUS₂, 0]

Status:

MINUS₁: multiset
s₁: [1]
IFMINUS₂: [2,1]
le: multiset
false: multiset
0: multiset
true: multiset

(18) Obligation:

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

MINUS(s0(x)) → IFMINUS(le, s0(x))
IFMINUS(false0, s0(x)) → MINUS(x)

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)
le → le
divL0(z, cons0(x, xs)) → divL0(z, consSwap0(x, xs))
le → false0
ifminus(s0(x)) → 00
div(s0(x)) → s0(div(minus(x)))
consSwap0(x, cons0(y, xs)) → cons0(y, consSwap0(x, xs))
minus(00) → 00
divL0(x, nil0) → x
minus(s0(x)) → ifminus(s0(x))
le → true0
ifminus(s0(x)) → s0(minus(x))

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

(19) MRRProof (EQUIVALENT transformation)

divL0(x, nil0) → x

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0
POL(IFMINUS(x₁, x₂)) = x₁ + 2·x₂
POL(MINUS(x₁)) = 2·x₁
POL(cons0(x₁, x₂)) = x₁ + x₂
POL(consSwap0(x₁, x₂)) = x₁ + x₂
POL(div(x₁)) = x₁
POL(divL0(x₁, x₂)) = 2 + 2·x₁ + 2·x₂
POL(false0) = 0
POL(ifminus(x₁)) = x₁
POL(le) = 0
POL(minus(x₁)) = x₁
POL(nil0) = 0
POL(s0(x₁)) = x₁
POL(true0) = 0

(20) Obligation:

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

MINUS(s0(x)) → IFMINUS(le, s0(x))
IFMINUS(false0, s0(x)) → MINUS(x)

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)
le → le
divL0(z, cons0(x, xs)) → divL0(z, consSwap0(x, xs))
le → false0
ifminus(s0(x)) → 00
div(s0(x)) → s0(div(minus(x)))
consSwap0(x, cons0(y, xs)) → cons0(y, consSwap0(x, xs))
minus(00) → 00
minus(s0(x)) → ifminus(s0(x))
le → true0
ifminus(s0(x)) → s0(minus(x))

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

(21) MRRProof (EQUIVALENT transformation)

IFMINUS(false0, s0(x)) → MINUS(x)

Strictly oriented rules of the TRS R:

le → true0

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 2
POL(IFMINUS(x₁, x₂)) = 2 + x₁ + x₂
POL(MINUS(x₁)) = 2 + 2·x₁
POL(cons0(x₁, x₂)) = x₁ + x₂
POL(consSwap0(x₁, x₂)) = x₁ + x₂
POL(div(x₁)) = x₁
POL(divL0(x₁, x₂)) = x₁ + 2·x₂
POL(false0) = 2
POL(ifminus(x₁)) = x₁
POL(le) = 2
POL(minus(x₁)) = x₁
POL(s0(x₁)) = 2 + 2·x₁
POL(true0) = 1

(22) Obligation:

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

MINUS(s0(x)) → IFMINUS(le, s0(x))

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)
le → le
divL0(z, cons0(x, xs)) → divL0(z, consSwap0(x, xs))
le → false0
ifminus(s0(x)) → 00
div(s0(x)) → s0(div(minus(x)))
consSwap0(x, cons0(y, xs)) → cons0(y, consSwap0(x, xs))
minus(00) → 00
minus(s0(x)) → ifminus(s0(x))
ifminus(s0(x)) → s0(minus(x))

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

(23) DependencyGraphProof (EQUIVALENT transformation)

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

(24) TRUE

(25) 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)
le(s(x), s(y)) → le(x, y)
divL(z, cons(x, xs)) → divL(z, consSwap(x, xs))
le(s(x), 0) → false
ifminus(true, s(x), y) → 0
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
minus(0, y) → 0
divL(x, nil) → x
minus(s(x), y) → ifminus(le(s(x), y), s(x), y)
le(0, y) → true
ifminus(false, s(x), y) → s(minus(x, y))

(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:true = ifminus_3 = 0, 2 divL_2 = false = nil = s_1 = div_2 = 1 0 = le_2 = 0, 1 minus_2 = 1 cons_2 = consSwap_2 = DIV_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.
DIV(x1, x2) = DIV(x1)
s(x1) = s(x1)
minus(x1, x2) = x1
0 = 0
ifminus(x1, x2, x3) = x2
le(x1, x2) = le
false = false
true = true
div(x1, x2) = x1

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

[DIV₁, s₁] > 0 > false
[le, true] > 0 > false

Status:

DIV₁: [1]
s₁: multiset
0: multiset
le: []
false: multiset
true: multiset

(27) 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
consSwap0(x, xs) → cons0(x, xs)
divL0(x, cons0(y, xs)) → divL0(div(x), xs)
le → le
divL0(z, cons0(x, xs)) → divL0(z, consSwap0(x, xs))
le → false0
ifminus(s0(x)) → 00
div(s0(x)) → s0(div(minus(x)))
consSwap0(x, cons0(y, xs)) → cons0(y, consSwap0(x, xs))
minus(00) → 00
divL0(x, nil0) → x
minus(s0(x)) → ifminus(s0(x))
le → true0
ifminus(s0(x)) → s0(minus(x))

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

(28) MRRProof (EQUIVALENT transformation)

le → false0
divL0(x, nil0) → x

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0
POL(DIV(x₁)) = 2·x₁
POL(cons0(x₁, x₂)) = x₁ + x₂
POL(consSwap0(x₁, x₂)) = x₁ + x₂
POL(div(x₁)) = x₁
POL(divL0(x₁, x₂)) = 2 + 2·x₁ + 2·x₂
POL(false0) = 0
POL(ifminus(x₁)) = x₁
POL(le) = 2
POL(minus(x₁)) = x₁
POL(nil0) = 0
POL(s0(x₁)) = 2·x₁
POL(true0) = 2

(29) 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
consSwap0(x, xs) → cons0(x, xs)
divL0(x, cons0(y, xs)) → divL0(div(x), xs)
le → le
divL0(z, cons0(x, xs)) → divL0(z, consSwap0(x, xs))
ifminus(s0(x)) → 00
div(s0(x)) → s0(div(minus(x)))
consSwap0(x, cons0(y, xs)) → cons0(y, consSwap0(x, xs))
minus(00) → 00
minus(s0(x)) → ifminus(s0(x))
le → true0
ifminus(s0(x)) → s0(minus(x))

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

(30) MRRProof (EQUIVALENT transformation)

le → true0

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0
POL(DIV(x₁)) = x₁
POL(cons0(x₁, x₂)) = x₁ + x₂
POL(consSwap0(x₁, x₂)) = x₁ + x₂
POL(div(x₁)) = x₁
POL(divL0(x₁, x₂)) = x₁ + 2·x₂
POL(ifminus(x₁)) = x₁
POL(le) = 2
POL(minus(x₁)) = x₁
POL(s0(x₁)) = x₁
POL(true0) = 1

(31) 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
consSwap0(x, xs) → cons0(x, xs)
divL0(x, cons0(y, xs)) → divL0(div(x), xs)
le → le
divL0(z, cons0(x, xs)) → divL0(z, consSwap0(x, xs))
ifminus(s0(x)) → 00
div(s0(x)) → s0(div(minus(x)))
consSwap0(x, cons0(y, xs)) → cons0(y, consSwap0(x, xs))
minus(00) → 00
minus(s0(x)) → ifminus(s0(x))
ifminus(s0(x)) → s0(minus(x))

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

(32) MRRProof (EQUIVALENT transformation)

DIV(s0(x)) → DIV(minus(x))

Strictly oriented rules of the TRS R:

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

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 2
POL(DIV(x₁)) = x₁
POL(cons0(x₁, x₂)) = 1 + x₁ + x₂
POL(consSwap0(x₁, x₂)) = 1 + x₁ + x₂
POL(div(x₁)) = x₁
POL(divL0(x₁, x₂)) = x₁ + 2·x₂
POL(ifminus(x₁)) = x₁
POL(le) = 0
POL(minus(x₁)) = x₁
POL(s0(x₁)) = 2 + 2·x₁

(33) Obligation:

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

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

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

(34) PisEmptyProof (EQUIVALENT transformation)

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

(35) YES

(36) 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)
le(s(x), s(y)) → le(x, y)
le(s(x), 0) → false
ifminus(true, s(x), y) → 0
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
divL(z, cons(x, xs)) → DIVL(z, CONSSWAP(x, xs))
minus(0, y) → 0
divL(x, nil) → x
minus(s(x), y) → ifminus(le(s(x), y), s(x), y)
le(0, y) → true
ifminus(false, s(x), y) → s(minus(x, y))

(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, 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)
le(s(x), s(y)) → le(x, y)
le(s(x), 0) → false
ifminus(true, s(x), y) → 0
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
minus(0, y) → 0
divL(x, nil) → x
minus(s(x), y) → ifminus(le(s(x), y), s(x), y)
le(0, y) → true
ifminus(false, s(x), y) → s(minus(x, y))

The remaining rules can at least be oriented weakly:

Ordered with Polynomial interpretation [POLO]:

POL(0) = 2
POL(CONSSWAP(x₁, x₂)) = 0
POL(DIV(x₁, x₂)) = 2 + 2·x₁·x₂
POL(DIVL(x₁, x₂)) = 2·x₁ + 3·x₂
POL(IFMINUS(x₁, x₂, x₃)) = 2 + x₁·x₂·x₃ + 2·x₃
POL(LE(x₁, x₂)) = 2
POL(MINUS(x₁, x₂)) = 2
POL(cons(x₁, x₂)) = 3 + 3·x₁ + x₂
POL(consSwap(x₁, x₂)) = 3 + 3·x₁ + x₂
POL(div(x₁, x₂)) = 1 + 3·x₂
POL(divL(x₁, x₂)) = x₁ + 2·x₂
POL(false) = 0
POL(ifminus(x₁, x₂, x₃)) = 3 + 3·x₁ + 3·x₂
POL(le(x₁, x₂)) = 0
POL(minus(x₁, x₂)) = 3 + 3·x₁
POL(nil) = 0
POL(s(x₁)) = 3
POL(true) = 0

(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)
le(s(x), s(y)) → le(x, y)
le(s(x), 0) → false
ifminus(true, s(x), y) → 0
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
divL(z, cons(x, xs)) → DIVL(z, CONSSWAP(x, xs))
minus(0, y) → 0
divL(x, nil) → x
minus(s(x), y) → ifminus(le(s(x), y), s(x), y)
le(0, y) → true
ifminus(false, s(x), y) → s(minus(x, y))

(39) DAbsisEmptyProof (EQUIVALENT transformation)

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

(40) YES

(41) 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)
le(s(x), s(y)) → le(x, y)
le(s(x), 0) → false
ifminus(true, s(x), y) → 0
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
divL(z, cons(x, xs)) → DIVL(z, CONSSWAP(x, xs))
minus(0, y) → 0
divL(x, nil) → x
minus(s(x), y) → ifminus(le(s(x), y), s(x), y)
le(0, y) → true
ifminus(false, s(x), y) → s(minus(x, y))

(42) 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)
le(s(x), s(y)) → le(x, y)
le(s(x), 0) → false
ifminus(true, s(x), y) → 0
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
minus(0, y) → 0
divL(x, nil) → x
minus(s(x), y) → ifminus(le(s(x), y), s(x), y)
le(0, y) → true
ifminus(false, s(x), y) → s(minus(x, y))

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 + 3·x₂
POL(IFMINUS(x₁, x₂, x₃)) = 2 + x₁·x₂ + 2·x₃
POL(LE(x₁, x₂)) = 2 + 2·x₁·x₂
POL(MINUS(x₁, x₂)) = 2 + 2·x₁
POL(cons(x₁, x₂)) = 2
POL(consSwap(x₁, x₂)) = 2
POL(div(x₁, x₂)) = x₁
POL(divL(x₁, x₂)) = x₁
POL(false) = 0
POL(ifminus(x₁, x₂, x₃)) = 3·x₁ + x₂
POL(le(x₁, x₂)) = 0
POL(minus(x₁, x₂)) = x₁
POL(nil) = 2
POL(s(x₁)) = 1 + 3·x₁
POL(true) = 0

(43) 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)
le(s(x), s(y)) → le(x, y)
le(s(x), 0) → false
ifminus(true, s(x), y) → 0
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
divL(z, cons(x, xs)) → DIVL(z, CONSSWAP(x, xs))
minus(0, y) → 0
divL(x, nil) → x
minus(s(x), y) → ifminus(le(s(x), y), s(x), y)
le(0, y) → true
ifminus(false, s(x), y) → s(minus(x, y))

(44) DAbsisEmptyProof (EQUIVALENT transformation)

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

(45) YES

(46) 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)
le(s(x), s(y)) → le(x, y)
le(s(x), 0) → false
ifminus(true, s(x), y) → 0
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
divL(z, cons(x, xs)) → DIVL(z, CONSSWAP(x, xs))
minus(0, y) → 0
divL(x, nil) → x
minus(s(x), y) → ifminus(le(s(x), y), s(x), y)
le(0, y) → true
ifminus(false, s(x), y) → s(minus(x, y))

(47) 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)
le(s(x), s(y)) → le(x, y)
le(s(x), 0) → false
ifminus(true, s(x), y) → 0
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
minus(0, y) → 0
divL(x, nil) → x
minus(s(x), y) → ifminus(le(s(x), y), s(x), y)
le(0, y) → true
ifminus(false, s(x), y) → s(minus(x, y))

The remaining rules can at least be oriented weakly:

Ordered with Polynomial interpretation [POLO]:

POL(0) = 2
POL(CONSSWAP(x₁, x₂)) = 0
POL(DIV(x₁, x₂)) = 2 + 2·x₁·x₂
POL(DIVL(x₁, x₂)) = 2·x₁ + 3·x₂
POL(IFMINUS(x₁, x₂, x₃)) = 2 + x₁·x₂·x₃ + 2·x₃
POL(LE(x₁, x₂)) = 2
POL(MINUS(x₁, x₂)) = 2
POL(cons(x₁, x₂)) = 3 + 3·x₁ + x₂
POL(consSwap(x₁, x₂)) = 3 + 3·x₁ + x₂
POL(div(x₁, x₂)) = 1 + 3·x₂
POL(divL(x₁, x₂)) = x₁ + 2·x₂
POL(false) = 0
POL(ifminus(x₁, x₂, x₃)) = 3 + 3·x₁ + 3·x₂
POL(le(x₁, x₂)) = 0
POL(minus(x₁, x₂)) = 3 + 3·x₁
POL(nil) = 0
POL(s(x₁)) = 3
POL(true) = 0

(48) 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)
le(s(x), s(y)) → le(x, y)
le(s(x), 0) → false
ifminus(true, s(x), y) → 0
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
divL(z, cons(x, xs)) → DIVL(z, CONSSWAP(x, xs))
minus(0, y) → 0
divL(x, nil) → x
minus(s(x), y) → ifminus(le(s(x), y), s(x), y)
le(0, y) → true
ifminus(false, s(x), y) → s(minus(x, y))

(49) DAbsisEmptyProof (EQUIVALENT transformation)

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

(50) YES

(51) 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)
le(s(x), s(y)) → le(x, y)
le(s(x), 0) → false
ifminus(true, s(x), y) → 0
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
divL(z, cons(x, xs)) → DIVL(z, CONSSWAP(x, xs))
minus(0, y) → 0
minus(s(x), y) → ifminus(le(s(x), y), s(x), y)
le(0, y) → true
ifminus(false, s(x), y) → s(minus(x, y))

(52) 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)
le(s(x), s(y)) → le(x, y)
le(s(x), 0) → false
ifminus(true, s(x), y) → 0
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
minus(s(x), y) → ifminus(le(s(x), y), s(x), y)
le(0, y) → true
ifminus(false, s(x), y) → s(minus(x, y))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
minus(0, y) → 0

The remaining rules can at least be oriented weakly:

Ordered with Polynomial interpretation [POLO]:

POL(0) = 2
POL(CONSSWAP(x₁, x₂)) = 0
POL(DIV(x₁, x₂)) = 2 + 2·x₁·x₂ + x₂
POL(DIVL(x₁, x₂)) = 1 + 3·x₁ + x₂
POL(IFMINUS(x₁, x₂, x₃)) = 2 + x₁·x₃ + 2·x₃
POL(LE(x₁, x₂)) = 2
POL(MINUS(x₁, x₂)) = 2·x₁ + x₁·x₂
POL(cons(x₁, x₂)) = 2 + 3·x₂
POL(consSwap(x₁, x₂)) = 2 + 3·x₂
POL(div(x₁, x₂)) = 3
POL(divL(x₁, x₂)) = 2 + x₁ + 2·x₂
POL(false) = 2
POL(ifminus(x₁, x₂, x₃)) = 3 + 3·x₂ + 2·x₃
POL(le(x₁, x₂)) = 2·x₁ + x₂
POL(minus(x₁, x₂)) = 3 + 3·x₁ + 2·x₂
POL(nil) = 0
POL(s(x₁)) = x₁
POL(true) = 0

(53) 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)
le(s(x), s(y)) → le(x, y)
le(s(x), 0) → false
ifminus(true, s(x), y) → 0
div(s(x), s(y)) → s(div(minus(x, y), s(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
minus(0, y) → 0
minus(s(x), y) → ifminus(le(s(x), y), s(x), y)
le(0, y) → true
ifminus(false, s(x), y) → s(minus(x, y))

(54) 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) MRRProof (EQUIVALENT transformation)

(13) Obligation:

(14) PisEmptyProof (EQUIVALENT transformation)

(15) YES

(16) Obligation:

(17) RelADPCleverAfsProof (SOUND transformation)

(18) Obligation:

(19) MRRProof (EQUIVALENT transformation)

(20) Obligation:

(21) MRRProof (EQUIVALENT transformation)

(22) Obligation:

(23) DependencyGraphProof (EQUIVALENT transformation)

(24) TRUE

(25) Obligation:

(26) RelADPCleverAfsProof (SOUND transformation)

(27) Obligation:

(28) MRRProof (EQUIVALENT transformation)

(29) Obligation:

(30) MRRProof (EQUIVALENT transformation)

(31) Obligation:

(32) MRRProof (EQUIVALENT transformation)

(33) Obligation:

(34) PisEmptyProof (EQUIVALENT transformation)

(35) YES

(36) Obligation:

(37) RelADPReductionPairProof (EQUIVALENT transformation)

(38) Obligation:

(39) DAbsisEmptyProof (EQUIVALENT transformation)

(40) YES

(41) Obligation:

(42) RelADPReductionPairProof (EQUIVALENT transformation)

(43) Obligation:

(44) DAbsisEmptyProof (EQUIVALENT transformation)

(45) YES

(46) Obligation:

(47) RelADPReductionPairProof (EQUIVALENT transformation)

(48) Obligation:

(49) DAbsisEmptyProof (EQUIVALENT transformation)

(50) YES

(51) Obligation:

(52) RelADPReductionPairProof (EQUIVALENT transformation)

(53) Obligation:

(54) DAbsisEmptyProof (EQUIVALENT transformation)

(55) YES