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 (⇒, 94 ms)

        ↳7 QDP

        ↳8 MRRProof (⇔, 0 ms)

        ↳9 QDP

        ↳10 PisEmptyProof (⇔, 0 ms)

        ↳11 YES

    ↳12 RelADPP

        ↳13 RelADPCleverAfsProof (⇒, 100 ms)

        ↳14 QDP

        ↳15 MRRProof (⇔, 0 ms)

        ↳16 QDP

        ↳17 PisEmptyProof (⇔, 0 ms)

        ↳18 YES

    ↳19 RelADPP

        ↳20 RelADPCleverAfsProof (⇒, 141 ms)

        ↳21 QDP

        ↳22 MRRProof (⇔, 0 ms)

        ↳23 QDP

        ↳24 PisEmptyProof (⇔, 0 ms)

        ↳25 YES

    ↳26 RelADPP

        ↳27 RelADPReductionPairProof (⇔, 58 ms)

        ↳28 RelADPP

        ↳29 DAbsisEmptyProof (⇔, 0 ms)

        ↳30 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:

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)
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:

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(x, s(y)) → pred(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))
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))
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:MINUS_2 = 0 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 =
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) = x2
s(x1) = s(x1)
gcd(x1, x2) = gcd(x1, x2)
if_gcd(x1, x2, x3) = if_gcd(x2, x3)
le(x1, x2) = le
false = false
minus(x1, x2) = x1
true = true
0 = 0
pred(x1) = x1

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

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

Status:

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

(7) Obligation:

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

MINUS(s0(y)) → MINUS(y)

The TRS R consists of the following rules:

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))
pred0(s0(x)) → x
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))
gcdL0(cons0(x, cons0(y, xs))) → gcdL0(cons0(y, cons0(x, 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 dependency pairs:

MINUS(s0(y)) → MINUS(y)

Strictly oriented rules of the TRS R:

gcdL0(cons0(x, nil0)) → x
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
le → true0

Used ordering: Polynomial interpretation [POLO]:

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

(9) Obligation:

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

gcdL0(nil0) → 00
le → le
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))
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))
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))
gcdL(cons(x, cons(y, xs))) → gcdL(cons(y, cons(x, xs)))
le(0, y) → true
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:true = gcd_2 = pred_1 = if_gcd_3 = 0 nil = false = s_1 = gcdL_1 = LE_2 = 0 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.
LE(x1, x2) = LE(x2)
s(x1) = s(x1)
if_gcd(x1, x2, x3) = if_gcd(x2, x3)
true = true
gcd(x1, x2) = gcd(x1, x2)
minus(x1, x2) = minus(x1)
false = false
le(x1, x2) = le
0 = 0
pred(x1) = x1

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

[s₁, minus₁] > LE₁
[s₁, minus₁] > false > [if_gcd₂, gcd₂]
le > true > [if_gcd₂, gcd₂]
le > false > [if_gcd₂, gcd₂]

Status:

LE₁: multiset
s₁: multiset
if_gcd₂: multiset
true: multiset
gcd₂: multiset
minus₁: multiset
false: multiset
le: []
0: multiset

(14) Obligation:

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

LE(s0(y)) → LE(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))
pred0(s0(x)) → x
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))
gcdL0(cons0(x, cons0(y, xs))) → gcdL0(cons0(y, cons0(x, xs)))
le → true0
minus(x) → x

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

(15) MRRProof (EQUIVALENT transformation)

LE(s0(y)) → LE(y)

Strictly oriented rules of the TRS R:

gcdL0(cons0(x, nil0)) → x
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
le → true0

Used ordering: Polynomial interpretation [POLO]:

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

(16) Obligation:

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

gcdL0(nil0) → 00
le → le
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))
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.

(17) PisEmptyProof (EQUIVALENT transformation)

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

(18) YES

(19) 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))
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))
gcdL(cons(x, cons(y, xs))) → gcdL(cons(y, cons(x, xs)))
le(0, y) → true
minus(x, 0) → x

(20) 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 =
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

(21) 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 → 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))
pred0(s0(x)) → x
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))
gcdL0(cons0(x, cons0(y, xs))) → gcdL0(cons0(y, cons0(x, xs)))
le → true0
minus(x) → x

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

(22) 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
gcd0(s0(x), 00) → s0(x)
pred0(s0(x)) → 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))
le → true0
minus(x) → x

Used ordering: Polynomial interpretation [POLO]:

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

(23) Obligation:

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

gcdL0(nil0) → 00
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))
minus(x) → pred0(minus(x))
gcdL0(cons0(x, cons0(y, xs))) → gcdL0(cons0(y, cons0(x, xs)))

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

(24) PisEmptyProof (EQUIVALENT transformation)

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

(25) YES

(26) 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))
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
gcdL(cons(x, cons(y, xs))) → GCDL(cons(y, cons(x, xs)))

(27) 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))
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) = 1
POL(GCD(x₁, x₂)) = 2 + x₁
POL(GCDL(x₁)) = 2 + 3·x₁
POL(IF_GCD(x₁, x₂, x₃)) = 0
POL(LE(x₁, x₂)) = x₂
POL(MINUS(x₁, x₂)) = 2·x₁
POL(PRED(x₁)) = 2
POL(cons(x₁, x₂)) = 1 + 3·x₁ + x₂
POL(false) = 0
POL(gcd(x₁, x₂)) = x₁ + x₂
POL(gcdL(x₁)) = 1 + 2·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(pred(x₁)) = x₁
POL(s(x₁)) = 2·x₁
POL(true) = 0

(28) 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))
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
gcdL(cons(x, cons(y, xs))) → GCDL(cons(y, cons(x, xs)))

(29) 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) PisEmptyProof (EQUIVALENT transformation)

(11) YES

(12) Obligation:

(13) RelADPCleverAfsProof (SOUND transformation)

(14) Obligation:

(15) MRRProof (EQUIVALENT transformation)

(16) Obligation:

(17) PisEmptyProof (EQUIVALENT transformation)

(18) YES

(19) Obligation:

(20) RelADPCleverAfsProof (SOUND transformation)

(21) Obligation:

(22) MRRProof (EQUIVALENT transformation)

(23) Obligation:

(24) PisEmptyProof (EQUIVALENT transformation)

(25) YES

(26) Obligation:

(27) RelADPReductionPairProof (EQUIVALENT transformation)

(28) Obligation:

(29) DAbsisEmptyProof (EQUIVALENT transformation)

(30) YES