YES

The TRS could be proven terminating. The proof took 3529 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (138ms).
 | – Problem 2 was processed with processor SubtermCriterion (0ms).
 | – Problem 3 was processed with processor SubtermCriterion (1ms).
 | – Problem 4 was processed with processor SubtermCriterion (0ms).
 | – Problem 5 was processed with processor PolynomialOrderingProcessor (2257ms).

Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

gcd#(s(x), s(y), z)min#(x, y)gcd#(s(x), s(y), z)max#(x, y)
gcd#(s(x), y, s(z))max#(x, z)max#(s(x), s(y))max#(x, y)
gcd#(s(x), y, s(z))gcd#(-(max(x, z), min(x, z)), y, s(min(x, z)))gcd#(s(x), y, s(z))-#(max(x, z), min(x, z))
gcd#(s(x), s(y), z)gcd#(-(max(x, y), min(x, y)), s(min(x, y)), z)-#(s(x), s(y))-#(x, y)
gcd#(x, s(y), s(z))min#(y, z)gcd#(x, s(y), s(z))max#(y, z)
gcd#(s(x), y, s(z))min#(x, z)min#(s(x), s(y))min#(x, y)
gcd#(x, s(y), s(z))-#(max(y, z), min(y, z))gcd#(s(x), s(y), z)-#(max(x, y), min(x, y))
gcd#(x, s(y), s(z))gcd#(x, -(max(y, z), min(y, z)), s(min(y, z)))

Rewrite Rules

min(x, 0)0min(0, y)0
min(s(x), s(y))s(min(x, y))max(x, 0)x
max(0, y)ymax(s(x), s(y))s(max(x, y))
-(x, 0)x-(s(x), s(y))-(x, y)
gcd(s(x), s(y), z)gcd(-(max(x, y), min(x, y)), s(min(x, y)), z)gcd(x, s(y), s(z))gcd(x, -(max(y, z), min(y, z)), s(min(y, z)))
gcd(s(x), y, s(z))gcd(-(max(x, z), min(x, z)), y, s(min(x, z)))gcd(x, 0, 0)x
gcd(0, y, 0)ygcd(0, 0, z)z

Original Signature

Termination of terms over the following signature is verified: min, 0, max, s, -, gcd

Strategy


The following SCCs where found

min#(s(x), s(y)) → min#(x, y)

max#(s(x), s(y)) → max#(x, y)

-#(s(x), s(y)) → -#(x, y)

gcd#(s(x), y, s(z)) → gcd#(-(max(x, z), min(x, z)), y, s(min(x, z)))gcd#(s(x), s(y), z) → gcd#(-(max(x, y), min(x, y)), s(min(x, y)), z)
gcd#(x, s(y), s(z)) → gcd#(x, -(max(y, z), min(y, z)), s(min(y, z)))

Problem 2: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

-#(s(x), s(y))-#(x, y)

Rewrite Rules

min(x, 0)0min(0, y)0
min(s(x), s(y))s(min(x, y))max(x, 0)x
max(0, y)ymax(s(x), s(y))s(max(x, y))
-(x, 0)x-(s(x), s(y))-(x, y)
gcd(s(x), s(y), z)gcd(-(max(x, y), min(x, y)), s(min(x, y)), z)gcd(x, s(y), s(z))gcd(x, -(max(y, z), min(y, z)), s(min(y, z)))
gcd(s(x), y, s(z))gcd(-(max(x, z), min(x, z)), y, s(min(x, z)))gcd(x, 0, 0)x
gcd(0, y, 0)ygcd(0, 0, z)z

Original Signature

Termination of terms over the following signature is verified: min, 0, max, s, -, gcd

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

-#(s(x), s(y))-#(x, y)

Problem 3: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

min#(s(x), s(y))min#(x, y)

Rewrite Rules

min(x, 0)0min(0, y)0
min(s(x), s(y))s(min(x, y))max(x, 0)x
max(0, y)ymax(s(x), s(y))s(max(x, y))
-(x, 0)x-(s(x), s(y))-(x, y)
gcd(s(x), s(y), z)gcd(-(max(x, y), min(x, y)), s(min(x, y)), z)gcd(x, s(y), s(z))gcd(x, -(max(y, z), min(y, z)), s(min(y, z)))
gcd(s(x), y, s(z))gcd(-(max(x, z), min(x, z)), y, s(min(x, z)))gcd(x, 0, 0)x
gcd(0, y, 0)ygcd(0, 0, z)z

Original Signature

Termination of terms over the following signature is verified: min, 0, max, s, -, gcd

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

min#(s(x), s(y))min#(x, y)

Problem 4: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

max#(s(x), s(y))max#(x, y)

Rewrite Rules

min(x, 0)0min(0, y)0
min(s(x), s(y))s(min(x, y))max(x, 0)x
max(0, y)ymax(s(x), s(y))s(max(x, y))
-(x, 0)x-(s(x), s(y))-(x, y)
gcd(s(x), s(y), z)gcd(-(max(x, y), min(x, y)), s(min(x, y)), z)gcd(x, s(y), s(z))gcd(x, -(max(y, z), min(y, z)), s(min(y, z)))
gcd(s(x), y, s(z))gcd(-(max(x, z), min(x, z)), y, s(min(x, z)))gcd(x, 0, 0)x
gcd(0, y, 0)ygcd(0, 0, z)z

Original Signature

Termination of terms over the following signature is verified: min, 0, max, s, -, gcd

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

max#(s(x), s(y))max#(x, y)

Problem 5: PolynomialOrderingProcessor



Dependency Pair Problem

Dependency Pairs

gcd#(s(x), y, s(z))gcd#(-(max(x, z), min(x, z)), y, s(min(x, z)))gcd#(s(x), s(y), z)gcd#(-(max(x, y), min(x, y)), s(min(x, y)), z)
gcd#(x, s(y), s(z))gcd#(x, -(max(y, z), min(y, z)), s(min(y, z)))

Rewrite Rules

min(x, 0)0min(0, y)0
min(s(x), s(y))s(min(x, y))max(x, 0)x
max(0, y)ymax(s(x), s(y))s(max(x, y))
-(x, 0)x-(s(x), s(y))-(x, y)
gcd(s(x), s(y), z)gcd(-(max(x, y), min(x, y)), s(min(x, y)), z)gcd(x, s(y), s(z))gcd(x, -(max(y, z), min(y, z)), s(min(y, z)))
gcd(s(x), y, s(z))gcd(-(max(x, z), min(x, z)), y, s(min(x, z)))gcd(x, 0, 0)x
gcd(0, y, 0)ygcd(0, 0, z)z

Original Signature

Termination of terms over the following signature is verified: min, 0, max, s, -, gcd

Strategy


Polynomial Interpretation

Improved Usable rules

-(s(x), s(y))-(x, y)min(0, y)0
-(x, 0)xmax(s(x), s(y))s(max(x, y))
min(s(x), s(y))s(min(x, y))max(0, y)y
min(x, 0)0max(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), y, s(z))gcd#(-(max(x, z), min(x, z)), y, s(min(x, z)))gcd#(s(x), s(y), z)gcd#(-(max(x, y), min(x, y)), s(min(x, y)), z)
gcd#(x, s(y), s(z))gcd#(x, -(max(y, z), min(y, z)), s(min(y, z)))