YES Termination proof of AProVE_24_gcdSet_b.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 (⇒, 109 ms)
        ↳7 QDP
        ↳8 MRRProof (⇔, 0 ms)
        ↳9 QDP
        ↳10 PisEmptyProof (⇔, 0 ms)
        ↳11 YES
    ↳12 RelADPP
        ↳13 RelADPCleverAfsProof (⇒, 87 ms)
        ↳14 QDP
        ↳15 MRRProof (⇔, 0 ms)
        ↳16 QDP
        ↳17 MRRProof (⇔, 0 ms)
        ↳18 QDP
        ↳19 PisEmptyProof (⇔, 0 ms)
        ↳20 YES
    ↳21 RelADPP
        ↳22 RelADPCleverAfsProof (⇒, 100 ms)
        ↳23 QDP
        ↳24 MRRProof (⇔, 0 ms)
        ↳25 QDP
        ↳26 PisEmptyProof (⇔, 0 ms)
        ↳27 YES
    ↳28 RelADPP
        ↳29 RelADPReductionPairProof (⇔, 108 ms)
        ↳30 RelADPP
        ↳31 DAbsisEmptyProof (⇔, 0 ms)
        ↳32 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(x, 0) → x
minus(s(x), s(y)) → 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:

gcdL(cons(x, cons(y, xs))) → gcdL(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(x, 0) → x
minus(s(x), s(y)) → 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:

gcdL(cons(x, cons(y, xs))) → GCDL(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:

minus(s(x), s(y)) → MINUS(x, y)

and relative ADPs:

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))
gcd(0, y) → y
gcd(s(x), s(y)) → if_gcd(le(y, x), s(x), s(y))
gcdL(cons(x, cons(y, xs))) → gcdL(cons(gcd(x, y), xs))
gcdL(cons(x, cons(y, xs))) → gcdL(cons(y, cons(x, 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:s_1 = MINUS_2 = gcdL_1 = true = gcd_2 = 0 = le_2 = 0, 1 minus_2 = 1 if_gcd_3 = 0 cons_2 = nil = false =
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)  =  x1
true  =  true
le(x1, x2)  =  le
0  =  0

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

[ifgcd2, false, gcd2, true, le, 0] > s1

Status:
MINUS2: [2,1]
s1: [1]
ifgcd2: multiset
false: multiset
gcd2: multiset
true: multiset
le: []
0: multiset

(7) Obligation:

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

MINUS0(s0(x), s0(y)) → MINUS0(x, y)

The TRS R consists of the following rules:

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))
gcd0(00, y) → y
gcd0(s0(x), s0(y)) → if_gcd(s0(x), s0(y))
gcdL0(cons0(x, cons0(y, xs))) → gcdL0(cons0(gcd0(x, y), xs))
gcdL0(cons0(x, cons0(y, xs))) → gcdL0(cons0(y, cons0(x, xs)))
le → true0
minus(s0(x)) → minus(x)
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 dependency pairs:

MINUS0(s0(x), s0(y)) → MINUS0(x, 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))
gcd0(00, y) → y
le → true0
minus(s0(x)) → minus(x)

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 2   
POL(MINUS0(x1, x2)) = 2·x1 + 2·x2   
POL(cons0(x1, x2)) = 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(nil0) = 1   
POL(s0(x1)) = 2 + x1   
POL(true0) = 0   

(9) Obligation:

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

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

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:

le(s(x), s(y)) → LE(x, y)

and relative ADPs:

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))
gcd(0, y) → y
gcd(s(x), s(y)) → if_gcd(le(y, x), s(x), s(y))
gcdL(cons(x, cons(y, xs))) → gcdL(cons(gcd(x, y), xs))
gcdL(cons(x, cons(y, xs))) → gcdL(cons(y, cons(x, xs)))
le(0, y) → true
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → 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 = gcdL_1 = true = LE_2 = gcd_2 = le_2 = 1 0 = if_gcd_3 = 0 minus_2 = 1 cons_2 = nil = false =
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)
gcd(x1, x2)  =  gcd(x1, x2)
if_gcd(x1, x2, x3)  =  if_gcd(x2, x3)
le(x1, x2)  =  le(x1)
true  =  true
minus(x1, x2)  =  minus(x1)
false  =  false
0  =  0

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

[gcd2, ifgcd2] > [s1, le1, false] > minus1 > LE2
0 > true > [s1, le1, false] > minus1 > LE2

Status:
LE2: [1,2]
s1: multiset
gcd2: multiset
ifgcd2: multiset
le1: [1]
true: multiset
minus1: [1]
false: multiset
0: multiset

(14) 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:

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

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

Used ordering: Polynomial interpretation [POLO]:

POL(00) = 1   
POL(LE0(x1, x2)) = x1 + x2   
POL(cons0(x1, x2)) = 2 + x1 + x2   
POL(false0) = 1   
POL(gcd0(x1, x2)) = 1 + x1 + x2   
POL(gcdL0(x1)) = 2 + x1   
POL(if_gcd(x1, x2)) = 1 + x1 + x2   
POL(le(x1)) = 2 + x1   
POL(minus(x1)) = x1   
POL(nil0) = 2   
POL(s0(x1)) = 2·x1   
POL(true0) = 0   

(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:

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

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:

LE0(s0(x), s0(y)) → LE0(x, y)

Strictly oriented rules of the TRS R:

le(s0(x)) → le(x)
if_gcd(s0(x), s0(y)) → gcd0(minus(x), s0(y))
if_gcd(s0(x), s0(y)) → gcd0(minus(y), s0(x))
minus(s0(x)) → minus(x)

Used ordering: Polynomial interpretation [POLO]:

POL(LE0(x1, x2)) = x1 + x2   
POL(cons0(x1, x2)) = 2·x1 + x2   
POL(gcd0(x1, x2)) = 2 + x1 + x2   
POL(gcdL0(x1)) = 2·x1   
POL(if_gcd(x1, x2)) = 2 + x1 + x2   
POL(le(x1)) = 2·x1   
POL(minus(x1)) = x1   
POL(s0(x1)) = 1 + 2·x1   

(18) Obligation:

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

gcd0(s0(x), s0(y)) → if_gcd(s0(x), s0(y))
gcdL0(cons0(x, cons0(y, xs))) → gcdL0(cons0(y, cons0(x, xs)))
minus(x) → x

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:

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))
gcd(0, y) → y
gcd(s(x), s(y)) → if_gcd(le(y, x), s(x), s(y))
gcdL(cons(x, cons(y, xs))) → gcdL(cons(gcd(x, y), xs))
gcdL(cons(x, cons(y, xs))) → gcdL(cons(y, cons(x, xs)))
le(0, y) → true
minus(s(x), s(y)) → minus(x, y)
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 = if_gcd_3 = 0 GCD_2 = false = nil = s_1 = gcdL_1 = le_2 = 0 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.
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(x2)
true  =  true
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:

[s1, le1] > [false, true, 0] > [ifgcd2, gcd2] > [IFGCD2, GCD2]

Status:
IFGCD2: multiset
false: multiset
s1: [1]
GCD2: multiset
le1: [1]
true: multiset
0: multiset
ifgcd2: 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:

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

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

(24) 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:

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

Used ordering: Polynomial interpretation [POLO]:

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

(25) Obligation:

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

gcd0(s0(x), s0(y)) → if_gcd(s0(x), s0(y))
gcdL0(cons0(x, cons0(y, xs))) → gcdL0(cons0(y, cons0(x, xs)))

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:

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))
gcd(0, y) → y
gcd(s(x), s(y)) → if_gcd(le(y, x), s(x), s(y))
gcdL(cons(x, cons(y, xs))) → gcdL(cons(gcd(x, y), xs))
le(0, y) → true
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
gcdL(cons(x, cons(y, xs))) → GCDL(cons(y, cons(x, xs)))

(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:

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


The remaining rules can at least be oriented weakly:

Ordered with Polynomial interpretation [POLO]:

POL(0) = 0   
POL(GCD(x1, x2)) = 2   
POL(GCDL(x1)) = 3 + 3·x1   
POL(IF_GCD(x1, x2, x3)) = 2   
POL(LE(x1, x2)) = 2   
POL(MINUS(x1, x2)) = 2   
POL(cons(x1, x2)) = 3 + 3·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)) = x1   
POL(nil) = 0   
POL(s(x1)) = 3 + x1   
POL(true) = 0   

(30) Obligation:

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

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))
gcd(0, y) → y
gcd(s(x), s(y)) → if_gcd(le(y, x), s(x), s(y))
gcdL(cons(x, cons(y, xs))) → gcdL(cons(gcd(x, y), xs))
le(0, y) → true
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
gcdL(cons(x, cons(y, xs))) → GCDL(cons(y, cons(x, xs)))

(31) DAbsisEmptyProof (EQUIVALENT transformation)

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

(32) YES