Termination proof of AProVE_24_gcdSet

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 (⇒, 93 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 (⇒, 103 ms)

        ↳16 QDP

        ↳17 MRRProof (⇔, 0 ms)

        ↳18 QDP

        ↳19 PisEmptyProof (⇔, 0 ms)

        ↳20 YES

    ↳21 RelADPP

        ↳22 RelADPCleverAfsProof (⇒, 106 ms)

        ↳23 QDP

        ↳24 MRRProof (⇔, 0 ms)

        ↳25 QDP

        ↳26 PisEmptyProof (⇔, 0 ms)

        ↳27 YES

    ↳28 RelADPP

        ↳29 RelADPReductionPairProof (⇔, 100 ms)

        ↳30 RelADPP

        ↳31 DAbsisEmptyProof (⇔, 0 ms)

        ↳32 YES

    ↳33 RelADPP

        ↳34 RelADPReductionPairProof (⇔, 122 ms)

        ↳35 RelADPP

        ↳36 DAbsisEmptyProof (⇔, 0 ms)

        ↳37 YES

    ↳38 RelADPP

        ↳39 RelADPReductionPairProof (⇔, 97 ms)

        ↳40 RelADPP

        ↳41 DAbsisEmptyProof (⇔, 0 ms)

        ↳42 YES

    ↳43 RelADPP

        ↳44 RelADPReductionPairProof (⇔, 108 ms)

        ↳45 RelADPP

        ↳46 DAbsisEmptyProof (⇔, 0 ms)

        ↳47 YES

    ↳48 RelADPP

        ↳49 RelADPReductionPairProof (⇔, 106 ms)

        ↳50 RelADPP

        ↳51 DAbsisEmptyProof (⇔, 0 ms)

        ↳52 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)
pred(s(x)) → x
minus(x, 0) → x
minus(x, s(y)) → pred(minus(x, y))
gcd(0, y) → y
gcd(s(x), 0) → s(x)
gcd(s(x), s(y)) → if_gcd(le(y, x), s(x), s(y))
if_gcd(true, s(x), s(y)) → gcd(minus(x, y), s(y))
if_gcd(false, s(x), s(y)) → gcd(minus(y, x), s(x))
gcdL(nil) → 0
gcdL(cons(x, nil)) → x
gcdL(cons(x, cons(y, xs))) → gcdL(cons(gcd(x, y), xs))

The relative TRS consists of the following S rules:

consSwap(x, xs) → cons(x, xs)
gcdL(cons(x, xs)) → gcdL(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)
pred(s(x)) → x
minus(x, 0) → x
minus(x, s(y)) → PRED(minus(x, y))
minus(x, s(y)) → pred(MINUS(x, y))
gcd(0, y) → y
gcd(s(x), 0) → s(x)
gcd(s(x), s(y)) → IF_GCD(le(y, x), s(x), s(y))
gcd(s(x), s(y)) → if_gcd(LE(y, x), s(x), s(y))
if_gcd(true, s(x), s(y)) → GCD(minus(x, y), s(y))
if_gcd(true, s(x), s(y)) → gcd(MINUS(x, y), s(y))
if_gcd(false, s(x), s(y)) → GCD(minus(y, x), s(x))
if_gcd(false, s(x), s(y)) → gcd(MINUS(y, x), s(x))
gcdL(nil) → 0
gcdL(cons(x, nil)) → x
gcdL(cons(x, cons(y, xs))) → GCDL(cons(gcd(x, y), xs))
gcdL(cons(x, cons(y, xs))) → gcdL(cons(GCD(x, y), xs))

and relative ADPs:

consSwap(x, xs) → cons(x, xs)
gcdL(cons(x, xs)) → GCDL(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,
4 Lassos,
Result: This relative DT problem is equivalent to 8 subproblems.

(4) Complex Obligation (AND)

(5) Obligation:

Relative ADP Problem with
absolute ADPs:

minus(x, s(y)) → pred(MINUS(x, y))

and relative ADPs:

consSwap(x, xs) → cons(x, xs)
gcdL(cons(x, nil)) → x
gcdL(nil) → 0
le(s(x), s(y)) → le(x, y)
le(s(x), 0) → false
gcd(s(x), 0) → s(x)
if_gcd(true, s(x), s(y)) → gcd(minus(x, y), s(y))
if_gcd(false, s(x), s(y)) → gcd(minus(y, x), s(x))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
pred(s(x)) → x
gcdL(cons(x, xs)) → gcdL(consSwap(x, xs))
gcd(0, y) → y
gcd(s(x), s(y)) → if_gcd(le(y, x), s(x), s(y))
minus(x, s(y)) → pred(minus(x, y))
gcdL(cons(x, cons(y, xs))) → gcdL(cons(gcd(x, y), xs))
le(0, y) → true
minus(x, 0) → 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:MINUS_2 = true = gcd_2 = pred_1 = if_gcd_3 = 0 nil = false = s_1 = gcdL_1 = 0 = le_2 = 0, 1 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, x2)
s(x1) = s(x1)
if_gcd(x1, x2, x3) = if_gcd(x2, x3)
false = false
gcd(x1, x2) = gcd(x1, x2)
minus(x1, x2) = minus(x1)
true = true
le(x1, x2) = le
0 = 0
pred(x1) = x1

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

[true, le] > [false, 0] > [if_gcd₂, gcd₂] > [s₁, minus₁] > MINUS₂

Status:

MINUS₂: [1,2]
s₁: multiset
if_gcd₂: multiset
false: multiset
gcd₂: multiset
minus₁: multiset
true: multiset
le: multiset
0: multiset

(7) Obligation:

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

MINUS0(x, s0(y)) → MINUS0(x, y)

The TRS R consists of the following rules:

consSwap0(x, xs) → cons0(x, xs)
gcdL0(cons0(x, nil0)) → x
gcdL0(nil0) → 00
le → le
le → false0
gcd0(s0(x), 00) → s0(x)
if_gcd(s0(x), s0(y)) → gcd0(minus(x), s0(y))
if_gcd(s0(x), s0(y)) → gcd0(minus(y), s0(x))
consSwap0(x, cons0(y, xs)) → cons0(y, consSwap0(x, xs))
pred0(s0(x)) → x
gcdL0(cons0(x, xs)) → gcdL0(consSwap0(x, xs))
gcd0(00, y) → y
gcd0(s0(x), s0(y)) → if_gcd(s0(x), s0(y))
minus(x) → pred0(minus(x))
gcdL0(cons0(x, cons0(y, xs))) → gcdL0(cons0(gcd0(x, y), xs))
le → true0
minus(x) → 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:

gcdL0(cons0(x, nil0)) → x
gcdL0(nil0) → 00
le → true0

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0
POL(MINUS0(x₁, x₂)) = x₁ + x₂
POL(cons0(x₁, x₂)) = 2·x₁ + x₂
POL(consSwap0(x₁, x₂)) = 2·x₁ + x₂
POL(false0) = 2
POL(gcd0(x₁, x₂)) = x₁ + x₂
POL(gcdL0(x₁)) = x₁
POL(if_gcd(x₁, x₂)) = x₁ + x₂
POL(le) = 2
POL(minus(x₁)) = x₁
POL(nil0) = 1
POL(pred0(x₁)) = x₁
POL(s0(x₁)) = x₁
POL(true0) = 1

(9) Obligation:

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

MINUS0(x, s0(y)) → MINUS0(x, y)

The TRS R consists of the following rules:

consSwap0(x, xs) → cons0(x, xs)
le → le
le → false0
gcd0(s0(x), 00) → s0(x)
if_gcd(s0(x), s0(y)) → gcd0(minus(x), s0(y))
if_gcd(s0(x), s0(y)) → gcd0(minus(y), s0(x))
consSwap0(x, cons0(y, xs)) → cons0(y, consSwap0(x, xs))
pred0(s0(x)) → x
gcdL0(cons0(x, xs)) → gcdL0(consSwap0(x, xs))
gcd0(00, y) → y
gcd0(s0(x), s0(y)) → if_gcd(s0(x), s0(y))
minus(x) → pred0(minus(x))
gcdL0(cons0(x, cons0(y, xs))) → gcdL0(cons0(gcd0(x, y), xs))
minus(x) → x

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

(10) MRRProof (EQUIVALENT transformation)

MINUS0(x, s0(y)) → MINUS0(x, y)

Strictly oriented rules of the TRS R:

le → false0
gcd0(s0(x), 00) → s0(x)
if_gcd(s0(x), s0(y)) → gcd0(minus(x), s0(y))
if_gcd(s0(x), s0(y)) → gcd0(minus(y), s0(x))
pred0(s0(x)) → x
gcd0(00, y) → y

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 2
POL(MINUS0(x₁, x₂)) = x₁ + x₂
POL(cons0(x₁, x₂)) = 2·x₁ + x₂
POL(consSwap0(x₁, x₂)) = 2·x₁ + x₂
POL(false0) = 1
POL(gcd0(x₁, x₂)) = x₁ + x₂
POL(gcdL0(x₁)) = 2·x₁
POL(if_gcd(x₁, x₂)) = x₁ + x₂
POL(le) = 2
POL(minus(x₁)) = x₁
POL(pred0(x₁)) = x₁
POL(s0(x₁)) = 2 + 2·x₁

(11) Obligation:

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

consSwap0(x, xs) → cons0(x, xs)
le → le
consSwap0(x, cons0(y, xs)) → cons0(y, consSwap0(x, xs))
gcdL0(cons0(x, xs)) → gcdL0(consSwap0(x, xs))
gcd0(s0(x), s0(y)) → if_gcd(s0(x), s0(y))
minus(x) → pred0(minus(x))
gcdL0(cons0(x, cons0(y, xs))) → gcdL0(cons0(gcd0(x, y), xs))
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:

le(s(x), s(y)) → LE(x, y)

and relative ADPs:

consSwap(x, xs) → cons(x, xs)
gcdL(cons(x, nil)) → x
gcdL(nil) → 0
le(s(x), 0) → false
gcd(s(x), 0) → s(x)
if_gcd(true, s(x), s(y)) → gcd(minus(x, y), s(y))
if_gcd(false, s(x), s(y)) → gcd(minus(y, x), s(x))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
pred(s(x)) → x
gcdL(cons(x, xs)) → gcdL(consSwap(x, xs))
gcd(0, y) → y
gcd(s(x), s(y)) → if_gcd(le(y, x), s(x), s(y))
minus(x, s(y)) → pred(minus(x, y))
gcdL(cons(x, cons(y, xs))) → gcdL(cons(gcd(x, y), xs))
le(0, y) → true
minus(x, 0) → 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:true = gcd_2 = pred_1 = if_gcd_3 = 0 nil = false = s_1 = gcdL_1 = LE_2 = 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, x2)
s(x1) = s(x1)
if_gcd(x1, x2, x3) = if_gcd(x2, x3)
false = false
gcd(x1, x2) = gcd(x1, x2)
minus(x1, x2) = minus(x1)
le(x1, x2) = le
true = true
0 = 0
pred(x1) = x1

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

[le, true] > [if_gcd₂, gcd₂] > [s₁, false, minus₁] > LE₂

Status:

LE₂: [2,1]
s₁: [1]
if_gcd₂: multiset
false: multiset
gcd₂: multiset
minus₁: [1]
le: []
true: multiset
0: multiset

(16) Obligation:

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

LE0(s0(x), s0(y)) → LE0(x, y)

The TRS R consists of the following rules:

consSwap0(x, xs) → cons0(x, xs)
gcdL0(cons0(x, nil0)) → x
gcdL0(nil0) → 00
le → le
le → false0
gcd0(s0(x), 00) → s0(x)
if_gcd(s0(x), s0(y)) → gcd0(minus(x), s0(y))
if_gcd(s0(x), s0(y)) → gcd0(minus(y), s0(x))
consSwap0(x, cons0(y, xs)) → cons0(y, consSwap0(x, xs))
pred0(s0(x)) → x
gcdL0(cons0(x, xs)) → gcdL0(consSwap0(x, xs))
gcd0(00, y) → y
gcd0(s0(x), s0(y)) → if_gcd(s0(x), s0(y))
minus(x) → pred0(minus(x))
gcdL0(cons0(x, cons0(y, xs))) → gcdL0(cons0(gcd0(x, y), xs))
le → true0
minus(x) → x

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

(17) MRRProof (EQUIVALENT transformation)

LE0(s0(x), s0(y)) → LE0(x, y)

Strictly oriented rules of the TRS R:

gcdL0(cons0(x, nil0)) → x
gcdL0(nil0) → 00
gcd0(s0(x), 00) → s0(x)
pred0(s0(x)) → x
gcd0(00, y) → y
le → true0
minus(x) → x

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 2
POL(LE0(x₁, x₂)) = x₁ + x₂
POL(cons0(x₁, x₂)) = 2 + 2·x₁ + x₂
POL(consSwap0(x₁, x₂)) = 2 + 2·x₁ + x₂
POL(false0) = 2
POL(gcd0(x₁, x₂)) = 1 + x₁ + x₂
POL(gcdL0(x₁)) = 2 + x₁
POL(if_gcd(x₁, x₂)) = 1 + x₁ + x₂
POL(le) = 2
POL(minus(x₁)) = 1 + x₁
POL(nil0) = 1
POL(pred0(x₁)) = x₁
POL(s0(x₁)) = 1 + 2·x₁
POL(true0) = 1

(18) Obligation:

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

consSwap0(x, xs) → cons0(x, xs)
le → le
le → false0
if_gcd(s0(x), s0(y)) → gcd0(minus(x), s0(y))
if_gcd(s0(x), s0(y)) → gcd0(minus(y), s0(x))
consSwap0(x, cons0(y, xs)) → cons0(y, consSwap0(x, xs))
gcdL0(cons0(x, xs)) → gcdL0(consSwap0(x, xs))
gcd0(s0(x), s0(y)) → if_gcd(s0(x), s0(y))
minus(x) → pred0(minus(x))
gcdL0(cons0(x, cons0(y, xs))) → gcdL0(cons0(gcd0(x, y), 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:

if_gcd(true, s(x), s(y)) → GCD(minus(x, y), s(y))
gcd(s(x), s(y)) → IF_GCD(le(y, x), s(x), s(y))
if_gcd(false, s(x), s(y)) → GCD(minus(y, x), s(x))

and relative ADPs:

consSwap(x, xs) → cons(x, xs)
gcdL(cons(x, nil)) → x
gcdL(nil) → 0
le(s(x), s(y)) → le(x, y)
le(s(x), 0) → false
gcd(s(x), 0) → s(x)
if_gcd(true, s(x), s(y)) → gcd(minus(x, y), s(y))
if_gcd(false, s(x), s(y)) → gcd(minus(y, x), s(x))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
pred(s(x)) → x
gcdL(cons(x, xs)) → gcdL(consSwap(x, xs))
gcd(0, y) → y
gcd(s(x), s(y)) → if_gcd(le(y, x), s(x), s(y))
minus(x, s(y)) → pred(minus(x, y))
gcdL(cons(x, cons(y, xs))) → gcdL(cons(gcd(x, y), xs))
le(0, y) → true
minus(x, 0) → x

(22) 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 = IF_GCD_3 = 0 gcd_2 = pred_1 = if_gcd_3 = 0 GCD_2 = false = nil = s_1 = gcdL_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.
IF_GCD(x1, x2, x3) = IF_GCD(x2, x3)
false = false
s(x1) = s(x1)
GCD(x1, x2) = GCD(x1, x2)
minus(x1, x2) = x1
le(x1, x2) = le
true = true
pred(x1) = x1
0 = 0
if_gcd(x1, x2, x3) = if_gcd(x2, x3)
gcd(x1, x2) = gcd(x1, x2)

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

[if_gcd₂, gcd₂] > s₁ > [IF_GCD₂, false, GCD₂, le] > true

Status:

IF_GCD₂: multiset
false: multiset
s₁: [1]
GCD₂: multiset
le: []
true: multiset
0: multiset
if_gcd₂: multiset
gcd₂: multiset

(23) Obligation:

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

IF_GCD(s0(x), s0(y)) → GCD0(minus(y), s0(x))
GCD0(s0(x), s0(y)) → IF_GCD(s0(x), s0(y))
IF_GCD(s0(x), s0(y)) → GCD0(minus(x), s0(y))

The TRS R consists of the following rules:

consSwap0(x, xs) → cons0(x, xs)
gcdL0(cons0(x, nil0)) → x
gcdL0(nil0) → 00
le → le
le → false0
gcd0(s0(x), 00) → s0(x)
if_gcd(s0(x), s0(y)) → gcd0(minus(x), s0(y))
if_gcd(s0(x), s0(y)) → gcd0(minus(y), s0(x))
consSwap0(x, cons0(y, xs)) → cons0(y, consSwap0(x, xs))
pred0(s0(x)) → x
gcdL0(cons0(x, xs)) → gcdL0(consSwap0(x, xs))
gcd0(00, y) → y
gcd0(s0(x), s0(y)) → if_gcd(s0(x), s0(y))
minus(x) → pred0(minus(x))
gcdL0(cons0(x, cons0(y, xs))) → gcdL0(cons0(gcd0(x, y), xs))
le → true0
minus(x) → x

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

(24) MRRProof (EQUIVALENT transformation)

IF_GCD(s0(x), s0(y)) → GCD0(minus(y), s0(x))
GCD0(s0(x), s0(y)) → IF_GCD(s0(x), s0(y))
IF_GCD(s0(x), s0(y)) → GCD0(minus(x), s0(y))

Strictly oriented rules of the TRS R:

gcdL0(cons0(x, nil0)) → x
gcdL0(nil0) → 00
le → false0
gcd0(s0(x), 00) → s0(x)
if_gcd(s0(x), s0(y)) → gcd0(minus(x), s0(y))
if_gcd(s0(x), s0(y)) → gcd0(minus(y), s0(x))
pred0(s0(x)) → x
gcd0(00, y) → y
gcd0(s0(x), s0(y)) → if_gcd(s0(x), s0(y))
le → true0

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 0
POL(GCD0(x₁, x₂)) = 2 + 2·x₁ + x₂
POL(IF_GCD(x₁, x₂)) = 2·x₁ + x₂
POL(cons0(x₁, x₂)) = 2 + 2·x₁ + x₂
POL(consSwap0(x₁, x₂)) = 2 + 2·x₁ + x₂
POL(false0) = 1
POL(gcd0(x₁, x₂)) = 1 + x₁ + x₂
POL(gcdL0(x₁)) = x₁
POL(if_gcd(x₁, x₂)) = x₁ + x₂
POL(le) = 2
POL(minus(x₁)) = x₁
POL(nil0) = 1
POL(pred0(x₁)) = x₁
POL(s0(x₁)) = 2 + 2·x₁
POL(true0) = 0

(25) Obligation:

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

consSwap0(x, xs) → cons0(x, xs)
le → le
consSwap0(x, cons0(y, xs)) → cons0(y, consSwap0(x, xs))
gcdL0(cons0(x, xs)) → gcdL0(consSwap0(x, xs))
minus(x) → pred0(minus(x))
gcdL0(cons0(x, cons0(y, xs))) → gcdL0(cons0(gcd0(x, y), xs))
minus(x) → x

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

(26) PisEmptyProof (EQUIVALENT transformation)

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

(27) YES

(28) Obligation:

Relative ADP Problem with
absolute ADPs:

gcdL(cons(x, cons(y, xs))) → GCDL(cons(gcd(x, y), xs))

and relative ADPs:

consSwap(x, xs) → cons(x, xs)
gcdL(cons(x, nil)) → x
gcdL(nil) → 0
le(s(x), s(y)) → le(x, y)
le(s(x), 0) → false
gcd(s(x), 0) → s(x)
gcdL(cons(x, xs)) → GCDL(CONSSWAP(x, xs))
if_gcd(true, s(x), s(y)) → gcd(minus(x, y), s(y))
if_gcd(false, s(x), s(y)) → gcd(minus(y, x), s(x))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
pred(s(x)) → x
gcd(0, y) → y
gcd(s(x), s(y)) → if_gcd(le(y, x), s(x), s(y))
minus(x, s(y)) → pred(minus(x, y))
gcdL(cons(x, cons(y, xs))) → gcdL(cons(gcd(x, y), xs))
le(0, y) → true
minus(x, 0) → x

(29) 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:

gcdL(cons(x, cons(y, xs))) → GCDL(cons(gcd(x, y), xs))

Relative ADPs:

consSwap(x, xs) → cons(x, xs)
gcdL(cons(x, nil)) → x
gcdL(nil) → 0
le(s(x), s(y)) → le(x, y)
le(s(x), 0) → false
gcd(s(x), 0) → s(x)
if_gcd(true, s(x), s(y)) → gcd(minus(x, y), s(y))
if_gcd(false, s(x), s(y)) → gcd(minus(y, x), s(x))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
pred(s(x)) → x
gcd(0, y) → y
gcd(s(x), s(y)) → if_gcd(le(y, x), s(x), s(y))
minus(x, s(y)) → pred(minus(x, y))
gcdL(cons(x, cons(y, xs))) → gcdL(cons(gcd(x, y), xs))
le(0, y) → true
minus(x, 0) → x

The remaining rules can at least be oriented weakly:

Ordered with Polynomial interpretation [POLO]:

POL(0) = 2
POL(CONSSWAP(x₁, x₂)) = 0
POL(GCD(x₁, x₂)) = 2
POL(GCDL(x₁)) = 1 + 3·x₁
POL(IF_GCD(x₁, x₂, x₃)) = 2
POL(LE(x₁, x₂)) = 2
POL(MINUS(x₁, x₂)) = 0
POL(PRED(x₁)) = 2 + 2·x₁ + x₁²
POL(cons(x₁, x₂)) = 1 + 2·x₁ + x₂
POL(consSwap(x₁, x₂)) = 1 + 2·x₁ + x₂
POL(false) = 0
POL(gcd(x₁, x₂)) = x₁ + x₂
POL(gcdL(x₁)) = 1 + 2·x₁
POL(if_gcd(x₁, x₂, x₃)) = x₂ + x₃
POL(le(x₁, x₂)) = 2·x₁
POL(minus(x₁, x₂)) = x₁
POL(nil) = 2
POL(pred(x₁)) = x₁
POL(s(x₁)) = x₁
POL(true) = 3

(30) Obligation:

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

consSwap(x, xs) → cons(x, xs)
gcdL(cons(x, nil)) → x
gcdL(nil) → 0
le(s(x), s(y)) → le(x, y)
le(s(x), 0) → false
gcd(s(x), 0) → s(x)
gcdL(cons(x, xs)) → GCDL(CONSSWAP(x, xs))
if_gcd(true, s(x), s(y)) → gcd(minus(x, y), s(y))
if_gcd(false, s(x), s(y)) → gcd(minus(y, x), s(x))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
pred(s(x)) → x
gcd(0, y) → y
gcd(s(x), s(y)) → if_gcd(le(y, x), s(x), s(y))
minus(x, s(y)) → pred(minus(x, y))
gcdL(cons(x, cons(y, xs))) → gcdL(cons(gcd(x, y), xs))
le(0, y) → true
minus(x, 0) → x

(31) DAbsisEmptyProof (EQUIVALENT transformation)

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

(32) YES

(33) Obligation:

Relative ADP Problem with
absolute ADPs:

gcdL(cons(x, cons(y, xs))) → GCDL(cons(gcd(x, y), xs))

and relative ADPs:

consSwap(x, xs) → cons(x, xs)
gcdL(cons(x, nil)) → x
gcdL(nil) → 0
le(s(x), s(y)) → le(x, y)
le(s(x), 0) → false
gcd(s(x), 0) → s(x)
gcdL(cons(x, xs)) → GCDL(CONSSWAP(x, xs))
if_gcd(true, s(x), s(y)) → gcd(minus(x, y), s(y))
if_gcd(false, s(x), s(y)) → gcd(minus(y, x), s(x))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
pred(s(x)) → x
gcd(0, y) → y
gcd(s(x), s(y)) → if_gcd(le(y, x), s(x), s(y))
minus(x, s(y)) → pred(minus(x, y))
gcdL(cons(x, cons(y, xs))) → gcdL(cons(gcd(x, y), xs))
le(0, y) → true
minus(x, 0) → x

(34) 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:

gcdL(cons(x, cons(y, xs))) → GCDL(cons(gcd(x, y), xs))

Relative ADPs:

consSwap(x, xs) → cons(x, xs)
gcdL(cons(x, nil)) → x
gcdL(nil) → 0
le(s(x), s(y)) → le(x, y)
le(s(x), 0) → false
gcd(s(x), 0) → s(x)
if_gcd(true, s(x), s(y)) → gcd(minus(x, y), s(y))
if_gcd(false, s(x), s(y)) → gcd(minus(y, x), s(x))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
pred(s(x)) → x
gcd(0, y) → y
gcd(s(x), s(y)) → if_gcd(le(y, x), s(x), s(y))
minus(x, s(y)) → pred(minus(x, y))
gcdL(cons(x, cons(y, xs))) → gcdL(cons(gcd(x, y), xs))
le(0, y) → true
minus(x, 0) → x

The remaining rules can at least be oriented weakly:

Ordered with Polynomial interpretation [POLO]:

POL(0) = 2
POL(CONSSWAP(x₁, x₂)) = 0
POL(GCD(x₁, x₂)) = 2
POL(GCDL(x₁)) = 1 + 3·x₁
POL(IF_GCD(x₁, x₂, x₃)) = 2
POL(LE(x₁, x₂)) = 2
POL(MINUS(x₁, x₂)) = 0
POL(PRED(x₁)) = 2 + 2·x₁ + x₁²
POL(cons(x₁, x₂)) = 1 + 2·x₁ + x₂
POL(consSwap(x₁, x₂)) = 1 + 2·x₁ + x₂
POL(false) = 0
POL(gcd(x₁, x₂)) = x₁ + x₂
POL(gcdL(x₁)) = 1 + 2·x₁
POL(if_gcd(x₁, x₂, x₃)) = x₂ + x₃
POL(le(x₁, x₂)) = 2·x₁
POL(minus(x₁, x₂)) = x₁
POL(nil) = 2
POL(pred(x₁)) = x₁
POL(s(x₁)) = x₁
POL(true) = 3

(35) Obligation:

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

consSwap(x, xs) → cons(x, xs)
gcdL(cons(x, nil)) → x
gcdL(nil) → 0
le(s(x), s(y)) → le(x, y)
le(s(x), 0) → false
gcd(s(x), 0) → s(x)
gcdL(cons(x, xs)) → GCDL(CONSSWAP(x, xs))
if_gcd(true, s(x), s(y)) → gcd(minus(x, y), s(y))
if_gcd(false, s(x), s(y)) → gcd(minus(y, x), s(x))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
pred(s(x)) → x
gcd(0, y) → y
gcd(s(x), s(y)) → if_gcd(le(y, x), s(x), s(y))
minus(x, s(y)) → pred(minus(x, y))
gcdL(cons(x, cons(y, xs))) → gcdL(cons(gcd(x, y), xs))
le(0, y) → true
minus(x, 0) → x

(36) DAbsisEmptyProof (EQUIVALENT transformation)

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

(37) YES

(38) Obligation:

Relative ADP Problem with
absolute ADPs:

gcdL(nil) → 0

and relative ADPs:

consSwap(x, xs) → cons(x, xs)
gcdL(cons(x, nil)) → x
le(s(x), s(y)) → le(x, y)
le(s(x), 0) → false
gcd(s(x), 0) → s(x)
gcdL(cons(x, xs)) → GCDL(CONSSWAP(x, xs))
if_gcd(true, s(x), s(y)) → gcd(minus(x, y), s(y))
if_gcd(false, s(x), s(y)) → gcd(minus(y, x), s(x))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
pred(s(x)) → x
gcd(0, y) → y
gcd(s(x), s(y)) → if_gcd(le(y, x), s(x), s(y))
minus(x, s(y)) → pred(minus(x, y))
gcdL(cons(x, cons(y, xs))) → gcdL(cons(gcd(x, y), xs))
le(0, y) → true
minus(x, 0) → 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:

gcdL(nil) → 0

Relative ADPs:

consSwap(x, xs) → cons(x, xs)
gcdL(cons(x, nil)) → x
le(s(x), s(y)) → le(x, y)
le(s(x), 0) → false
gcd(s(x), 0) → s(x)
if_gcd(true, s(x), s(y)) → gcd(minus(x, y), s(y))
if_gcd(false, s(x), s(y)) → gcd(minus(y, x), s(x))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
pred(s(x)) → x
gcd(0, y) → y
gcd(s(x), s(y)) → if_gcd(le(y, x), s(x), s(y))
minus(x, s(y)) → pred(minus(x, y))
gcdL(cons(x, cons(y, xs))) → gcdL(cons(gcd(x, y), xs))
le(0, y) → true
minus(x, 0) → x

The remaining rules can at least be oriented weakly:

Ordered with Polynomial interpretation [POLO]:

POL(0) = 0
POL(CONSSWAP(x₁, x₂)) = 0
POL(GCD(x₁, x₂)) = 2
POL(GCDL(x₁)) = 1 + 3·x₁
POL(IF_GCD(x₁, x₂, x₃)) = 0
POL(LE(x₁, x₂)) = 2
POL(MINUS(x₁, x₂)) = 2·x₂
POL(PRED(x₁)) = 2 + x₁²
POL(cons(x₁, x₂)) = 1 + x₁ + x₂
POL(consSwap(x₁, x₂)) = 1 + x₁ + x₂
POL(false) = 0
POL(gcd(x₁, x₂)) = x₁ + x₂
POL(gcdL(x₁)) = x₁
POL(if_gcd(x₁, x₂, x₃)) = x₁ + x₂ + x₃
POL(le(x₁, x₂)) = 0
POL(minus(x₁, x₂)) = x₁
POL(nil) = 0
POL(pred(x₁)) = x₁
POL(s(x₁)) = x₁
POL(true) = 0

(40) Obligation:

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

consSwap(x, xs) → cons(x, xs)
gcdL(cons(x, nil)) → x
gcdL(nil) → 0
le(s(x), s(y)) → le(x, y)
le(s(x), 0) → false
gcd(s(x), 0) → s(x)
gcdL(cons(x, xs)) → GCDL(CONSSWAP(x, xs))
if_gcd(true, s(x), s(y)) → gcd(minus(x, y), s(y))
if_gcd(false, s(x), s(y)) → gcd(minus(y, x), s(x))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
pred(s(x)) → x
gcd(0, y) → y
gcd(s(x), s(y)) → if_gcd(le(y, x), s(x), s(y))
minus(x, s(y)) → pred(minus(x, y))
gcdL(cons(x, cons(y, xs))) → gcdL(cons(gcd(x, y), xs))
le(0, y) → true
minus(x, 0) → x

(41) DAbsisEmptyProof (EQUIVALENT transformation)

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

(42) YES

(43) Obligation:

Relative ADP Problem with
absolute ADPs:

gcdL(cons(x, cons(y, xs))) → gcdL(cons(GCD(x, y), xs))

and relative ADPs:

consSwap(x, xs) → cons(x, xs)
gcdL(cons(x, nil)) → x
gcdL(nil) → 0
le(s(x), s(y)) → le(x, y)
le(s(x), 0) → false
gcd(s(x), 0) → s(x)
gcdL(cons(x, xs)) → GCDL(CONSSWAP(x, xs))
if_gcd(true, s(x), s(y)) → gcd(minus(x, y), s(y))
if_gcd(false, s(x), s(y)) → gcd(minus(y, x), s(x))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
pred(s(x)) → x
gcd(0, y) → y
gcd(s(x), s(y)) → if_gcd(le(y, x), s(x), s(y))
minus(x, s(y)) → pred(minus(x, y))
gcdL(cons(x, cons(y, xs))) → gcdL(cons(gcd(x, y), xs))
le(0, y) → true
minus(x, 0) → x

(44) 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:

gcdL(cons(x, cons(y, xs))) → gcdL(cons(GCD(x, y), xs))

Relative ADPs:

consSwap(x, xs) → cons(x, xs)
gcdL(cons(x, nil)) → x
gcdL(nil) → 0
le(s(x), s(y)) → le(x, y)
le(s(x), 0) → false
gcd(s(x), 0) → s(x)
if_gcd(true, s(x), s(y)) → gcd(minus(x, y), s(y))
if_gcd(false, s(x), s(y)) → gcd(minus(y, x), s(x))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
pred(s(x)) → x
gcd(0, y) → y
gcd(s(x), s(y)) → if_gcd(le(y, x), s(x), s(y))
minus(x, s(y)) → pred(minus(x, y))
gcdL(cons(x, cons(y, xs))) → gcdL(cons(gcd(x, y), xs))
le(0, y) → true
minus(x, 0) → x

The remaining rules can at least be oriented weakly:

Ordered with Polynomial interpretation [POLO]:

POL(0) = 0
POL(CONSSWAP(x₁, x₂)) = 0
POL(GCD(x₁, x₂)) = 1 + 2·x₂
POL(GCDL(x₁)) = 2 + 3·x₁
POL(IF_GCD(x₁, x₂, x₃)) = 2
POL(LE(x₁, x₂)) = 2
POL(MINUS(x₁, x₂)) = 2
POL(PRED(x₁)) = 2 + x₁
POL(cons(x₁, x₂)) = 1 + x₁ + x₂
POL(consSwap(x₁, x₂)) = 1 + x₁ + x₂
POL(false) = 0
POL(gcd(x₁, x₂)) = x₁ + x₂
POL(gcdL(x₁)) = x₁
POL(if_gcd(x₁, x₂, x₃)) = 3·x₁ + x₂ + x₃
POL(le(x₁, x₂)) = 0
POL(minus(x₁, x₂)) = 2 + x₁
POL(nil) = 0
POL(pred(x₁)) = x₁
POL(s(x₁)) = 2 + x₁
POL(true) = 0

(45) Obligation:

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

consSwap(x, xs) → cons(x, xs)
gcdL(cons(x, nil)) → x
gcdL(nil) → 0
le(s(x), s(y)) → le(x, y)
le(s(x), 0) → false
gcd(s(x), 0) → s(x)
gcdL(cons(x, xs)) → GCDL(CONSSWAP(x, xs))
if_gcd(true, s(x), s(y)) → gcd(minus(x, y), s(y))
if_gcd(false, s(x), s(y)) → gcd(minus(y, x), s(x))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
pred(s(x)) → x
gcd(0, y) → y
gcd(s(x), s(y)) → if_gcd(le(y, x), s(x), s(y))
minus(x, s(y)) → pred(minus(x, y))
gcdL(cons(x, cons(y, xs))) → gcdL(cons(gcd(x, y), xs))
le(0, y) → true
minus(x, 0) → x

(46) DAbsisEmptyProof (EQUIVALENT transformation)

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

(47) YES

(48) Obligation:

Relative ADP Problem with
absolute ADPs:

gcdL(cons(x, nil)) → x

and relative ADPs:

consSwap(x, xs) → cons(x, xs)
gcdL(nil) → 0
le(s(x), s(y)) → le(x, y)
le(s(x), 0) → false
gcd(s(x), 0) → s(x)
gcdL(cons(x, xs)) → GCDL(CONSSWAP(x, xs))
if_gcd(true, s(x), s(y)) → gcd(minus(x, y), s(y))
if_gcd(false, s(x), s(y)) → gcd(minus(y, x), s(x))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
pred(s(x)) → x
gcd(0, y) → y
gcd(s(x), s(y)) → if_gcd(le(y, x), s(x), s(y))
minus(x, s(y)) → pred(minus(x, y))
gcdL(cons(x, cons(y, xs))) → gcdL(cons(gcd(x, y), xs))
le(0, y) → true
minus(x, 0) → x

(49) 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:

gcdL(cons(x, nil)) → x

Relative ADPs:

consSwap(x, xs) → cons(x, xs)
gcdL(nil) → 0
le(s(x), s(y)) → le(x, y)
le(s(x), 0) → false
gcd(s(x), 0) → s(x)
if_gcd(true, s(x), s(y)) → gcd(minus(x, y), s(y))
if_gcd(false, s(x), s(y)) → gcd(minus(y, x), s(x))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
pred(s(x)) → x
gcd(0, y) → y
gcd(s(x), s(y)) → if_gcd(le(y, x), s(x), s(y))
minus(x, s(y)) → pred(minus(x, y))
gcdL(cons(x, cons(y, xs))) → gcdL(cons(gcd(x, y), xs))
le(0, y) → true
minus(x, 0) → x

The remaining rules can at least be oriented weakly:

Ordered with Polynomial interpretation [POLO]:

POL(0) = 0
POL(CONSSWAP(x₁, x₂)) = 0
POL(GCD(x₁, x₂)) = 2
POL(GCDL(x₁)) = 3·x₁
POL(IF_GCD(x₁, x₂, x₃)) = 2
POL(LE(x₁, x₂)) = 2
POL(MINUS(x₁, x₂)) = 2 + 2·x₁ + 2·x₁·x₂ + 2·x₂
POL(PRED(x₁)) = 0
POL(cons(x₁, x₂)) = 1 + 2·x₁ + x₂
POL(consSwap(x₁, x₂)) = 1 + 2·x₁ + x₂
POL(false) = 0
POL(gcd(x₁, x₂)) = x₁ + x₂
POL(gcdL(x₁)) = x₁
POL(if_gcd(x₁, x₂, x₃)) = x₁ + x₂ + x₃
POL(le(x₁, x₂)) = 0
POL(minus(x₁, x₂)) = x₁
POL(nil) = 1
POL(pred(x₁)) = x₁
POL(s(x₁)) = x₁
POL(true) = 0

(50) Obligation:

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

consSwap(x, xs) → cons(x, xs)
gcdL(cons(x, nil)) → x
gcdL(nil) → 0
le(s(x), s(y)) → le(x, y)
le(s(x), 0) → false
gcd(s(x), 0) → s(x)
gcdL(cons(x, xs)) → GCDL(CONSSWAP(x, xs))
if_gcd(true, s(x), s(y)) → gcd(minus(x, y), s(y))
if_gcd(false, s(x), s(y)) → gcd(minus(y, x), s(x))
consSwap(x, cons(y, xs)) → cons(y, consSwap(x, xs))
pred(s(x)) → x
gcd(0, y) → y
gcd(s(x), s(y)) → if_gcd(le(y, x), s(x), s(y))
minus(x, s(y)) → pred(minus(x, y))
gcdL(cons(x, cons(y, xs))) → gcdL(cons(gcd(x, y), xs))
le(0, y) → true
minus(x, 0) → x

(51) 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) PisEmptyProof (EQUIVALENT transformation)

(20) YES

(21) Obligation:

(22) RelADPCleverAfsProof (SOUND transformation)

(23) Obligation:

(24) MRRProof (EQUIVALENT transformation)

(25) Obligation:

(26) PisEmptyProof (EQUIVALENT transformation)

(27) YES

(28) Obligation:

(29) RelADPReductionPairProof (EQUIVALENT transformation)

(30) Obligation:

(31) DAbsisEmptyProof (EQUIVALENT transformation)

(32) YES

(33) Obligation:

(34) RelADPReductionPairProof (EQUIVALENT transformation)

(35) Obligation:

(36) DAbsisEmptyProof (EQUIVALENT transformation)

(37) YES

(38) Obligation:

(39) RelADPReductionPairProof (EQUIVALENT transformation)

(40) Obligation:

(41) DAbsisEmptyProof (EQUIVALENT transformation)

(42) YES

(43) Obligation:

(44) RelADPReductionPairProof (EQUIVALENT transformation)

(45) Obligation:

(46) DAbsisEmptyProof (EQUIVALENT transformation)

(47) YES

(48) Obligation:

(49) RelADPReductionPairProof (EQUIVALENT transformation)

(50) Obligation:

(51) DAbsisEmptyProof (EQUIVALENT transformation)

(52) YES