YES

The TRS could be proven terminating. The proof took 843 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (36ms).
 | – Problem 2 was processed with processor SubtermCriterion (3ms).
 | – Problem 3 was processed with processor PolynomialLinearRange4iUR (282ms).
 |    | – Problem 5 was processed with processor PolynomialLinearRange4iUR (152ms).
 |    |    | – Problem 6 was processed with processor DependencyGraph (16ms).
 | – Problem 4 was processed with processor SubtermCriterion (1ms).

Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

le(0, y) → truele(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))

Original Signature

Termination of terms over the following signature is verified: 0, minus, le, s, if_gcd, true, false, gcd

Strategy


The following SCCs where found

if_gcd#(false, s(x), s(y)) → gcd#(minus(y, x), s(x))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))

le#(s(x), s(y)) → le#(x, y)

minus#(s(x), s(y)) → minus#(x, y)

Problem 2: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

le#(s(x), s(y)) → le#(x, y)

Rewrite Rules

le(0, y) → truele(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))

Original Signature

Termination of terms over the following signature is verified: 0, minus, le, s, if_gcd, true, false, gcd

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

le#(s(x), s(y)) → le#(x, y)

Problem 3: PolynomialLinearRange4iUR



Dependency Pair Problem

Dependency Pairs

if_gcd#(false, s(x), s(y)) → gcd#(minus(y, x), s(x))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))

Rewrite Rules

le(0, y) → truele(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))

Original Signature

Termination of terms over the following signature is verified: 0, minus, le, s, if_gcd, true, false, gcd

Strategy


Polynomial Interpretation

Improved Usable rules

minus(s(x), s(y)) → minus(x, y)le(s(x), s(y)) → le(x, y)
le(s(x), 0) → falsele(0, y) → true
minus(x, 0) → x

The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:

if_gcd#(false, s(x), s(y)) → gcd#(minus(y, x), s(x))

Problem 5: PolynomialLinearRange4iUR



Dependency Pair Problem

Dependency Pairs

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))

Rewrite Rules

le(0, y) → truele(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))

Original Signature

Termination of terms over the following signature is verified: minus, 0, s, le, if_gcd, false, true, gcd

Strategy


Polynomial Interpretation

Improved Usable rules

minus(s(x), s(y)) → minus(x, y)minus(x, 0) → x

The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:

gcd#(s(x), s(y)) → if_gcd#(le(y, x), s(x), s(y))

Problem 6: DependencyGraph



Dependency Pair Problem

Dependency Pairs

if_gcd#(true, s(x), s(y)) → gcd#(minus(x, y), s(y))

Rewrite Rules

le(0, y) → truele(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))

Original Signature

Termination of terms over the following signature is verified: 0, minus, le, s, if_gcd, true, false, gcd

Strategy


There are no SCCs!

Problem 4: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

minus#(s(x), s(y)) → minus#(x, y)

Rewrite Rules

le(0, y) → truele(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))

Original Signature

Termination of terms over the following signature is verified: 0, minus, le, s, if_gcd, true, false, gcd

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

minus#(s(x), s(y)) → minus#(x, y)