Termination proof of AProVE_24_conf

Termination of the given RelTRS could be proven:

0 RelTRS

↳1 RelTRSSCleanerProof (⇒, 0 ms)

↳2 RelTRS

↳3 RelTRStoRelADPProof (⇔, 0 ms)

↳4 RelADPP

↳5 RelADPDepGraphProof (⇔, 0 ms)

↳6 AND

    ↳7 RelADPP

        ↳8 RelADPReductionPairProof (⇔, 60 ms)

        ↳9 RelADPP

        ↳10 DAbsisEmptyProof (⇔, 0 ms)

        ↳11 YES

    ↳12 RelADPP

        ↳13 RelADPCleverAfsProof (⇒, 58 ms)

        ↳14 QDP

        ↳15 MRRProof (⇔, 4 ms)

        ↳16 QDP

        ↳17 DependencyGraphProof (⇔, 0 ms)

        ↳18 TRUE

(0) Obligation:

Relative term rewrite system:
The relative TRS consists of the following R rules:

le(s(s(x)), s(s(y))) → le(s(x), s(y))
le(s(s(x)), s(s(y))) → le(x, y)

The relative TRS consists of the following S rules:

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))
le(s(x), s(y)) → le(x, y)
le(s(x), 0) → false
gcd(s(x), 0) → s(x)
le(0, y) → true
minus(x, 0) → 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))
if_gcd(error, x, y) → if_gcd(error, 0, 0)
pred(s(x)) → x

(1) RelTRSSCleanerProof (SOUND transformation)

We have moved all duplicating rules from S to R → Duplicating rules:

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))

(2) Obligation:

Relative term rewrite system:
The relative TRS consists of the following R rules:

le(s(s(x)), s(s(y))) → le(s(x), s(y))
le(s(s(x)), s(s(y))) → le(x, 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(false, s(x), s(y)) → gcd(minus(y, x), s(x))

The relative TRS consists of the following S rules:

gcd(0, y) → y
minus(x, s(y)) → pred(minus(x, y))
le(s(x), s(y)) → le(x, y)
le(s(x), 0) → false
gcd(s(x), 0) → s(x)
le(0, y) → true
minus(x, 0) → x
if_gcd(error, x, y) → if_gcd(error, 0, 0)
pred(s(x)) → x

(3) RelTRStoRelADPProof (EQUIVALENT transformation)

We upgrade the RelTRS problem to an equivalent Relative ADP Problem [IJCAR24].

(4) Obligation:

Relative ADP Problem with
absolute ADPs:

le(s(s(x)), s(s(y))) → LE(s(x), s(y))
le(s(s(x)), s(s(y))) → LE(x, y)
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))

and relative ADPs:

gcd(0, y) → y
minus(x, s(y)) → PRED(MINUS(x, y))
le(s(x), s(y)) → LE(x, y)
le(s(x), 0) → false
gcd(s(x), 0) → s(x)
le(0, y) → true
minus(x, 0) → x
if_gcd(error, x, y) → IF_GCD(error, 0, 0)
pred(s(x)) → x

(5) RelADPDepGraphProof (EQUIVALENT transformation)

We use the relative dependency graph processor [IJCAR24].
The approximation of the Relative Dependency Graph contains:
2 SCCs with nodes from P_abs,
0 Lassos,
Result: This relative DT problem is equivalent to 2 subproblems.

(6) Complex Obligation (AND)

(7) Obligation:

Relative ADP Problem with
absolute ADPs:

le(s(s(x)), s(s(y))) → LE(s(x), s(y))
le(s(s(x)), s(s(y))) → LE(x, y)

and relative ADPs:

gcd(s(x), s(y)) → if_gcd(le(y, x), s(x), s(y))
gcd(0, y) → y
minus(x, s(y)) → pred(minus(x, y))
le(s(x), 0) → false
gcd(s(x), 0) → s(x)
le(0, y) → true
if_gcd(true, s(x), s(y)) → gcd(minus(x, y), s(y))
minus(x, 0) → x
if_gcd(false, s(x), s(y)) → gcd(minus(y, x), s(x))
if_gcd(error, x, y) → if_gcd(error, 0, 0)
le(s(x), s(y)) → LE(x, y)
pred(s(x)) → x

(8) 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:

le(s(s(x)), s(s(y))) → LE(s(x), s(y))
le(s(s(x)), s(s(y))) → LE(x, y)

Relative ADPs:

gcd(s(x), s(y)) → if_gcd(le(y, x), s(x), s(y))
gcd(0, y) → y
minus(x, s(y)) → pred(minus(x, y))
le(s(x), 0) → false
gcd(s(x), 0) → s(x)
le(0, y) → true
if_gcd(true, s(x), s(y)) → gcd(minus(x, y), s(y))
minus(x, 0) → x
if_gcd(false, s(x), s(y)) → gcd(minus(y, x), s(x))
if_gcd(error, x, y) → if_gcd(error, 0, 0)
le(s(x), s(y)) → LE(x, y)
pred(s(x)) → x

No rules with annotations remain.
Ordered with Polynomial interpretation [POLO]:

POL(0) = 0
POL(GCD(x₁, x₂)) = 2
POL(IF_GCD(x₁, x₂, x₃)) = 2
POL(LE(x₁, x₂)) = 2 + x₁
POL(MINUS(x₁, x₂)) = 2·x₁ + 2·x₁·x₂
POL(PRED(x₁)) = 2
POL(error) = 0
POL(false) = 0
POL(gcd(x₁, x₂)) = 2 + 3·x₁ + 2·x₂
POL(if_gcd(x₁, x₂, x₃)) = 2 + 3·x₁ + 2·x₂ + 2·x₃
POL(le(x₁, x₂)) = 0
POL(minus(x₁, x₂)) = x₁
POL(pred(x₁)) = x₁
POL(s(x₁)) = 1 + 2·x₁
POL(true) = 0

(9) Obligation:

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

le(s(s(x)), s(s(y))) → le(s(x), s(y))
le(s(x), s(y)) → le(x, y)
le(s(s(x)), s(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))
if_gcd(error, x, y) → if_gcd(error, 0, 0)
pred(s(x)) → x
gcd(s(x), s(y)) → if_gcd(le(y, x), s(x), s(y))
gcd(0, y) → y
minus(x, s(y)) → pred(minus(x, y))
le(0, y) → true
minus(x, 0) → x

(10) DAbsisEmptyProof (EQUIVALENT transformation)

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

(11) YES

(12) 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:

le(s(s(x)), s(s(y))) → le(s(x), s(y))
le(s(x), s(y)) → le(x, y)
le(s(s(x)), s(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))
if_gcd(error, x, y) → if_gcd(error, 0, 0)
pred(s(x)) → x
gcd(s(x), s(y)) → if_gcd(le(y, x), s(x), s(y))
gcd(0, y) → y
minus(x, s(y)) → pred(minus(x, y))
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:s_1 = true = IF_GCD_3 = 0 error = gcd_2 = le_2 = 0, 1 0 = pred_1 = if_gcd_3 = 0 minus_2 = 1 GCD_2 = 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.
IF_GCD(x1, x2, x3) = IF_GCD(x2, x3)
false = false
s(x1) = s(x1)
GCD(x1, x2) = GCD(x1, x2)
minus(x1, x2) = minus(x1)
le(x1, x2) = le
true = true
pred(x1) = x1
0 = 0
gcd(x1, x2) = gcd(x1, x2)
if_gcd(x1, x2, x3) = if_gcd(x2, x3)
error = error

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

[IF_GCD₂, GCD₂] > s₁ > [false, minus₁, le, true] > [gcd₂, if_gcd₂] > error > 0

Status:

IF_GCD₂: multiset
false: multiset
s₁: [1]
GCD₂: multiset
minus₁: [1]
le: multiset
true: multiset
0: multiset
gcd₂: multiset
if_gcd₂: multiset
error: multiset

(14) 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:

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))
if_gcd(x, y) → if_gcd(00, 00)
pred0(s0(x)) → x
gcd0(s0(x), s0(y)) → if_gcd(s0(x), s0(y))
gcd0(00, y) → y
minus(x) → pred0(minus(x))
le → true0
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 dependency pairs:

GCD0(s0(x), s0(y)) → IF_GCD(s0(x), s0(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
le → true0
minus(x) → x

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

(16) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(17) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.

(0) Obligation:

(1) RelTRSSCleanerProof (SOUND transformation)

(2) Obligation:

(3) RelTRStoRelADPProof (EQUIVALENT transformation)

(4) Obligation:

(5) RelADPDepGraphProof (EQUIVALENT transformation)

(6) Complex Obligation (AND)

(7) Obligation:

(8) RelADPReductionPairProof (EQUIVALENT transformation)

(9) Obligation:

(10) DAbsisEmptyProof (EQUIVALENT transformation)

(11) YES

(12) Obligation:

(13) RelADPCleverAfsProof (SOUND transformation)

(14) Obligation:

(15) MRRProof (EQUIVALENT transformation)

(16) Obligation:

(17) DependencyGraphProof (EQUIVALENT transformation)

(18) TRUE