YES Termination proof of AProVE_24_mset_c.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 (⇒, 54 ms)
        ↳7 QDP
        ↳8 MRRProof (⇔, 0 ms)
        ↳9 QDP
        ↳10 MRRProof (⇔, 0 ms)
        ↳11 QDP
        ↳12 MRRProof (⇔, 0 ms)
        ↳13 QDP
        ↳14 PisEmptyProof (⇔, 0 ms)
        ↳15 YES
    ↳16 RelADPP
        ↳17 RelADPCleverAfsProof (⇒, 50 ms)
        ↳18 QDP
        ↳19 MRRProof (⇔, 0 ms)
        ↳20 QDP
        ↳21 MRRProof (⇔, 0 ms)
        ↳22 QDP
        ↳23 MRRProof (⇔, 0 ms)
        ↳24 QDP
        ↳25 PisEmptyProof (⇔, 0 ms)
        ↳26 YES
    ↳27 RelADPP
        ↳28 RelADPCleverAfsProof (⇒, 49 ms)
        ↳29 QDP
        ↳30 MRRProof (⇔, 0 ms)
        ↳31 QDP
        ↳32 MRRProof (⇔, 0 ms)
        ↳33 QDP
        ↳34 MRRProof (⇔, 0 ms)
        ↳35 QDP
        ↳36 PisEmptyProof (⇔, 0 ms)
        ↳37 YES
    ↳38 RelADPP
        ↳39 RelADPReductionPairProof (⇔, 59 ms)
        ↳40 RelADPP
        ↳41 DAbsisEmptyProof (⇔, 0 ms)
        ↳42 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:

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:

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:

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:
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:

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)))
le(s(x), 0) → false
minus(s(x), y) → ifminus(le(s(x), y), s(x), y)
ifminus(true, s(x), y) → 0
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
le(0, y) → true
ifminus(false, s(x), y) → s(minus(x, y))
minus(0, y) → 0
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 = true = ifminus_3 = 0, 2 div_2 = 1 LE_2 = 1 divL_2 = le_2 = 0, 1 0 = minus_2 = 1 cons_2 = false = 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.
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
divL0(x, cons0(y, xs)) → divL0(div(x), xs)
divL0(z, cons0(x, cons0(y, xs))) → divL0(z, cons0(y, cons0(x, xs)))
le → le
le → false0
minus(s0(x)) → ifminus(s0(x))
ifminus(s0(x)) → 00
div(s0(x)) → s0(div(minus(x)))
le → true0
ifminus(s0(x)) → s0(minus(x))
minus(00) → 00
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)
le → true0
divL0(x, nil0) → x

Used ordering: Polynomial interpretation [POLO]:

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

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

le → false0

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0   
POL(LE(x1)) = x1   
POL(cons0(x1, x2)) = x1 + x2   
POL(div(x1)) = x1   
POL(divL0(x1, x2)) = x1 + 2·x2   
POL(false0) = 1   
POL(ifminus(x1)) = x1   
POL(le) = 2   
POL(minus(x1)) = x1   
POL(s0(x1)) = 2·x1   

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

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

(12) 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:

LE(s0(x)) → LE(x)

Strictly oriented rules of the TRS R:

ifminus(s0(x)) → 00
minus(00) → 00

Used ordering: Polynomial interpretation [POLO]:

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

(13) 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)))
le → le
minus(s0(x)) → ifminus(s0(x))
div(s0(x)) → s0(div(minus(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
divL(x, cons(y, xs)) → divL(div(x, y), xs)
divL(z, cons(x, cons(y, xs))) → divL(z, cons(y, cons(x, xs)))
le(s(x), s(y)) → le(x, y)
le(s(x), 0) → false
minus(s(x), y) → ifminus(le(s(x), y), s(x), y)
ifminus(true, s(x), y) → 0
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
le(0, y) → true
minus(0, y) → 0
divL(x, nil) → x

(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 = 0, 2 false = nil = s_1 = div_2 = 1 le_2 = 0, 1 0 = minus_2 = 1 cons_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(x2)
le(x1, x2)  =  le
false  =  false
0  =  0
true  =  true
div(x1, x2)  =  x1
minus(x1, x2)  =  x1
ifminus(x1, x2, x3)  =  x2

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

[s1, le, true] > [false, 0] > [MINUS1, IFMINUS1]

Status:
MINUS1: multiset
s1: multiset
IFMINUS1: multiset
le: []
false: multiset
0: multiset
true: multiset

(18) Obligation:

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

MINUS(s0(x)) → IFMINUS(s0(x))
IFMINUS(s0(x)) → 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)))
le → le
le → false0
minus(s0(x)) → ifminus(s0(x))
ifminus(s0(x)) → 00
div(s0(x)) → s0(div(minus(x)))
le → true0
ifminus(s0(x)) → s0(minus(x))
minus(00) → 00
divL0(x, nil0) → 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:

divL0(x, cons0(y, xs)) → divL0(div(x), xs)
le → false0
divL0(x, nil0) → x

Used ordering: Polynomial interpretation [POLO]:

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

(20) Obligation:

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

MINUS(s0(x)) → IFMINUS(s0(x))
IFMINUS(s0(x)) → 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)))
le → le
minus(s0(x)) → ifminus(s0(x))
ifminus(s0(x)) → 00
div(s0(x)) → s0(div(minus(x)))
le → true0
ifminus(s0(x)) → s0(minus(x))
minus(00) → 00

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 → true0

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0   
POL(IFMINUS(x1)) = x1   
POL(MINUS(x1)) = x1   
POL(cons0(x1, x2)) = x1 + x2   
POL(div(x1)) = 2·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   

(22) Obligation:

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

MINUS(s0(x)) → IFMINUS(s0(x))
IFMINUS(s0(x)) → 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)))
le → le
minus(s0(x)) → ifminus(s0(x))
ifminus(s0(x)) → 00
div(s0(x)) → s0(div(minus(x)))
ifminus(s0(x)) → s0(minus(x))
minus(00) → 00

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

(23) 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)) → IFMINUS(s0(x))
IFMINUS(s0(x)) → MINUS(x)


Used ordering: Polynomial interpretation [POLO]:

POL(00) = 2   
POL(IFMINUS(x1)) = x1   
POL(MINUS(x1)) = 1 + 2·x1   
POL(cons0(x1, x2)) = 2·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   

(24) 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)))
le → le
minus(s0(x)) → ifminus(s0(x))
ifminus(s0(x)) → 00
div(s0(x)) → s0(div(minus(x)))
ifminus(s0(x)) → s0(minus(x))
minus(00) → 00

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

(25) PisEmptyProof (EQUIVALENT transformation)

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

(26) YES

(27) 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)))
le(s(x), s(y)) → le(x, y)
le(s(x), 0) → false
minus(s(x), y) → ifminus(le(s(x), y), s(x), y)
ifminus(true, s(x), y) → 0
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
le(0, y) → true
ifminus(false, s(x), y) → s(minus(x, y))
minus(0, y) → 0
divL(x, nil) → 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 = ifminus_3 = 0, 2 div_2 = 1 divL_2 = 0 = le_2 = 0, 1 minus_2 = 1 cons_2 = DIV_2 = 1 false = 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
ifminus(x1, x2, x3)  =  x2
le(x1, x2)  =  le
0  =  0
false  =  false
true  =  true
div(x1, x2)  =  div(x1)

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

[le, true] > [DIV1, s1, false] > 0
div1 > [DIV1, s1, false] > 0

Status:
DIV1: [1]
s1: multiset
le: []
0: multiset
false: multiset
true: multiset
div1: multiset

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

divL0(x, cons0(y, xs)) → divL0(div(x), xs)
le → true0
divL0(x, nil0) → x

Used ordering: Polynomial interpretation [POLO]:

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

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

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(DIV(x1)) = x1   
POL(cons0(x1, x2)) = x1 + x2   
POL(div(x1)) = x1   
POL(divL0(x1, x2)) = x1 + 2·x2   
POL(false0) = 1   
POL(ifminus(x1)) = x1   
POL(le) = 2   
POL(minus(x1)) = x1   
POL(s0(x1)) = 2·x1   

(33) 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)))
le → le
minus(s0(x)) → ifminus(s0(x))
ifminus(s0(x)) → 00
div(s0(x)) → s0(div(minus(x)))
ifminus(s0(x)) → s0(minus(x))
minus(00) → 00

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 dependency pairs:

DIV(s0(x)) → DIV(minus(x))

Strictly oriented rules of the TRS R:

div(00) → 00
ifminus(s0(x)) → 00
minus(00) → 00

Used ordering: Polynomial interpretation [POLO]:

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

(35) Obligation:

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

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

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

(36) PisEmptyProof (EQUIVALENT transformation)

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

(37) YES

(38) 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)
le(s(x), s(y)) → le(x, y)
le(s(x), 0) → false
minus(s(x), y) → ifminus(le(s(x), y), s(x), y)
ifminus(true, s(x), y) → 0
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
le(0, y) → true
ifminus(false, s(x), y) → s(minus(x, y))
divL(z, cons(x, cons(y, xs))) → DIVL(z, cons(y, cons(x, xs)))
minus(0, y) → 0
divL(x, nil) → x

(39) 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)
le(s(x), s(y)) → le(x, y)
le(s(x), 0) → false
minus(s(x), y) → ifminus(le(s(x), y), s(x), y)
ifminus(true, s(x), y) → 0
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
le(0, y) → true
ifminus(false, s(x), y) → s(minus(x, y))
minus(0, y) → 0
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)) = 2 + 3·x1 + 3·x2   
POL(IFMINUS(x1, x2, x3)) = 2·x1 + x1·x2·x3 + 2·x3   
POL(LE(x1, x2)) = x2   
POL(MINUS(x1, x2)) = x1·x2   
POL(cons(x1, x2)) = 3 + 3·x1 + x2   
POL(div(x1, x2)) = 2 + 2·x2   
POL(divL(x1, x2)) = 2 + 2·x1 + 2·x2   
POL(false) = 0   
POL(ifminus(x1, x2, x3)) = 3·x1 + 3·x2   
POL(le(x1, x2)) = 0   
POL(minus(x1, x2)) = 3·x1   
POL(nil) = 0   
POL(s(x1)) = x1   
POL(true) = 0   

(40) 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)
le(s(x), s(y)) → le(x, y)
le(s(x), 0) → false
minus(s(x), y) → ifminus(le(s(x), y), s(x), y)
ifminus(true, s(x), y) → 0
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
le(0, y) → true
ifminus(false, s(x), y) → s(minus(x, y))
divL(z, cons(x, cons(y, xs))) → DIVL(z, cons(y, cons(x, xs)))
minus(0, y) → 0
divL(x, nil) → x

(41) DAbsisEmptyProof (EQUIVALENT transformation)

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

(42) YES