YES

The TRS could be proven terminating. The proof took 659 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (41ms).
 | – Problem 2 was processed with processor SubtermCriterion (1ms).
 | – Problem 3 was processed with processor SubtermCriterion (1ms).
 | – Problem 4 was processed with processor PolynomialLinearRange4iUR (308ms).
 |    | – Problem 5 was processed with processor PolynomialLinearRange4iUR (153ms).
 |    |    | – Problem 6 was processed with processor DependencyGraph (15ms).

Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

if#(true, s(X), s(Y))minus#(X, Y)minus#(X, s(Y))pred#(minus(X, Y))
gcd#(s(X), s(Y))le#(Y, X)if#(true, s(X), s(Y))gcd#(minus(X, Y), s(Y))
if#(false, s(X), s(Y))gcd#(minus(Y, X), s(X))gcd#(s(X), s(Y))if#(le(Y, X), s(X), s(Y))
if#(false, s(X), s(Y))minus#(Y, X)minus#(X, s(Y))minus#(X, Y)
le#(s(X), s(Y))le#(X, Y)

Rewrite Rules

minus(X, s(Y))pred(minus(X, Y))minus(X, 0)X
pred(s(X))Xle(s(X), s(Y))le(X, Y)
le(s(X), 0)falsele(0, Y)true
gcd(0, Y)0gcd(s(X), 0)s(X)
gcd(s(X), s(Y))if(le(Y, X), s(X), s(Y))if(true, s(X), s(Y))gcd(minus(X, Y), s(Y))
if(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, false, true, pred, gcd

Strategy


The following SCCs where found

if#(true, s(X), s(Y)) → gcd#(minus(X, Y), s(Y))if#(false, s(X), s(Y)) → gcd#(minus(Y, X), s(X))
gcd#(s(X), s(Y)) → if#(le(Y, X), s(X), s(Y))

minus#(X, s(Y)) → minus#(X, Y)

le#(s(X), s(Y)) → le#(X, Y)

Problem 2: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

le#(s(X), s(Y))le#(X, Y)

Rewrite Rules

minus(X, s(Y))pred(minus(X, Y))minus(X, 0)X
pred(s(X))Xle(s(X), s(Y))le(X, Y)
le(s(X), 0)falsele(0, Y)true
gcd(0, Y)0gcd(s(X), 0)s(X)
gcd(s(X), s(Y))if(le(Y, X), s(X), s(Y))if(true, s(X), s(Y))gcd(minus(X, Y), s(Y))
if(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, false, true, pred, 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: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

minus#(X, s(Y))minus#(X, Y)

Rewrite Rules

minus(X, s(Y))pred(minus(X, Y))minus(X, 0)X
pred(s(X))Xle(s(X), s(Y))le(X, Y)
le(s(X), 0)falsele(0, Y)true
gcd(0, Y)0gcd(s(X), 0)s(X)
gcd(s(X), s(Y))if(le(Y, X), s(X), s(Y))if(true, s(X), s(Y))gcd(minus(X, Y), s(Y))
if(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, false, true, pred, gcd

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

minus#(X, s(Y))minus#(X, Y)

Problem 4: PolynomialLinearRange4iUR



Dependency Pair Problem

Dependency Pairs

if#(true, s(X), s(Y))gcd#(minus(X, Y), s(Y))if#(false, s(X), s(Y))gcd#(minus(Y, X), s(X))
gcd#(s(X), s(Y))if#(le(Y, X), s(X), s(Y))

Rewrite Rules

minus(X, s(Y))pred(minus(X, Y))minus(X, 0)X
pred(s(X))Xle(s(X), s(Y))le(X, Y)
le(s(X), 0)falsele(0, Y)true
gcd(0, Y)0gcd(s(X), 0)s(X)
gcd(s(X), s(Y))if(le(Y, X), s(X), s(Y))if(true, s(X), s(Y))gcd(minus(X, Y), s(Y))
if(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, false, true, pred, gcd

Strategy


Polynomial Interpretation

Improved Usable rules

pred(s(X))Xminus(X, 0)X
le(s(X), s(Y))le(X, Y)minus(X, s(Y))pred(minus(X, Y))
le(0, Y)truele(s(X), 0)false

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

if#(false, s(X), s(Y))gcd#(minus(Y, X), s(X))

Problem 5: PolynomialLinearRange4iUR



Dependency Pair Problem

Dependency Pairs

if#(true, s(X), s(Y))gcd#(minus(X, Y), s(Y))gcd#(s(X), s(Y))if#(le(Y, X), s(X), s(Y))

Rewrite Rules

minus(X, s(Y))pred(minus(X, Y))minus(X, 0)X
pred(s(X))Xle(s(X), s(Y))le(X, Y)
le(s(X), 0)falsele(0, Y)true
gcd(0, Y)0gcd(s(X), 0)s(X)
gcd(s(X), s(Y))if(le(Y, X), s(X), s(Y))if(true, s(X), s(Y))gcd(minus(X, Y), s(Y))
if(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, true, false, pred, gcd

Strategy


Polynomial Interpretation

Improved Usable rules

pred(s(X))Xminus(X, 0)X
minus(X, s(Y))pred(minus(X, Y))

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

if#(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#(le(Y, X), s(X), s(Y))

Rewrite Rules

minus(X, s(Y))pred(minus(X, Y))minus(X, 0)X
pred(s(X))Xle(s(X), s(Y))le(X, Y)
le(s(X), 0)falsele(0, Y)true
gcd(0, Y)0gcd(s(X), 0)s(X)
gcd(s(X), s(Y))if(le(Y, X), s(X), s(Y))if(true, s(X), s(Y))gcd(minus(X, Y), s(Y))
if(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, false, true, pred, gcd

Strategy


There are no SCCs!