YES

The TRS could be proven terminating. The proof took 842 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (59ms).
 | – Problem 2 was processed with processor SubtermCriterion (1ms).
 |    | – Problem 5 was processed with processor DependencyGraph (0ms).
 | – Problem 3 was processed with processor PolynomialLinearRange4iUR (606ms).
 |    | – Problem 6 was processed with processor DependencyGraph (1ms).
 | – 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_minus#(false, s(x), y)minus#(x, y)
if_gcd#(false, s(x), s(y))gcd#(minus(y, x), s(x))le#(s(x), s(y))le#(x, y)
minus#(s(x), y)le#(s(x), y)if_gcd#(true, s(x), s(y))minus#(x, y)
minus#(s(x), y)if_minus#(le(s(x), y), s(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(0, y)0
minus(s(x), y)if_minus(le(s(x), y), s(x), y)if_minus(true, s(x), y)0
if_minus(false, s(x), y)s(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: if_minus, 0, minus, le, s, if_gcd, true, false, gcd

Strategy


The following SCCs where found

if_minus#(false, s(x), y) → minus#(x, y)minus#(s(x), y) → if_minus#(le(s(x), y), s(x), y)

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)

Problem 2: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

if_minus#(false, s(x), y)minus#(x, y)minus#(s(x), y)if_minus#(le(s(x), y), s(x), y)

Rewrite Rules

le(0, y)truele(s(x), 0)false
le(s(x), s(y))le(x, y)minus(0, y)0
minus(s(x), y)if_minus(le(s(x), y), s(x), y)if_minus(true, s(x), y)0
if_minus(false, s(x), y)s(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: if_minus, 0, minus, le, s, if_gcd, true, false, gcd

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

if_minus#(false, s(x), y)minus#(x, y)

Problem 5: DependencyGraph



Dependency Pair Problem

Dependency Pairs

minus#(s(x), y)if_minus#(le(s(x), y), s(x), y)

Rewrite Rules

le(0, y)truele(s(x), 0)false
le(s(x), s(y))le(x, y)minus(0, y)0
minus(s(x), y)if_minus(le(s(x), y), s(x), y)if_minus(true, s(x), y)0
if_minus(false, s(x), y)s(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: if_minus, minus, 0, s, le, if_gcd, false, true, gcd

Strategy


There are no SCCs!

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(0, y)0
minus(s(x), y)if_minus(le(s(x), y), s(x), y)if_minus(true, s(x), y)0
if_minus(false, s(x), y)s(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: if_minus, 0, minus, le, s, if_gcd, true, false, gcd

Strategy


Polynomial Interpretation

Improved Usable rules

minus(0, y)0minus(s(x), y)if_minus(le(s(x), y), s(x), y)
if_minus(true, s(x), y)0if_minus(false, s(x), y)s(minus(x, y))

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

Problem 6: DependencyGraph



Dependency Pair Problem

Dependency Pairs

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(0, y)0
minus(s(x), y)if_minus(le(s(x), y), s(x), y)if_minus(true, s(x), y)0
if_minus(false, s(x), y)s(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: if_minus, minus, 0, s, le, if_gcd, false, true, gcd

Strategy


There are no SCCs!

Problem 4: 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(0, y)0
minus(s(x), y)if_minus(le(s(x), y), s(x), y)if_minus(true, s(x), y)0
if_minus(false, s(x), y)s(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: if_minus, 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)