TIMEOUT

The TRS could not be proven terminating. The proof attempt took 60013 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (44ms).
 | – Problem 2 was processed with processor SubtermCriterion (2ms).
 | – Problem 3 remains open; application of the following processors failed [SubtermCriterion (0ms), DependencyGraph (3ms), PolynomialLinearRange4iUR (431ms), DependencyGraph (4ms), PolynomialLinearRange8NegiUR (6848ms), DependencyGraph (27ms), ReductionPairSAT (533ms), DependencyGraph (5ms), SizeChangePrinciple (111ms), ForwardNarrowing (1ms), BackwardInstantiation (3ms), ForwardInstantiation (2ms), Propagation (0ms)].
 | – Problem 4 was processed with processor SubtermCriterion (1ms).

The following open problems remain:



Open Dependency Pair Problem 3

Dependency Pairs

if_gcd#(true, x, y)gcd#(minus(x, y), y)if_gcd#(false, x, y)gcd#(minus(y, x), x)
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(0, x)0minus(s(x), s(y))minus(x, y)
gcd(0, y)ygcd(s(x), 0)s(x)
gcd(s(x), s(y))if_gcd(le(y, x), s(x), s(y))if_gcd(true, x, y)gcd(minus(x, y), y)
if_gcd(false, x, y)gcd(minus(y, x), x)

Original Signature

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


Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

gcd#(s(x), s(y))le#(y, x)if_gcd#(true, x, y)gcd#(minus(x, y), y)
le#(s(x), s(y))le#(x, y)if_gcd#(false, x, y)minus#(y, x)
minus#(s(x), s(y))minus#(x, y)if_gcd#(true, x, y)minus#(x, y)
if_gcd#(false, x, y)gcd#(minus(y, x), x)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(0, x)0minus(s(x), s(y))minus(x, y)
gcd(0, y)ygcd(s(x), 0)s(x)
gcd(s(x), s(y))if_gcd(le(y, x), s(x), s(y))if_gcd(true, x, y)gcd(minus(x, y), y)
if_gcd(false, x, y)gcd(minus(y, x), 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

le#(s(x), s(y)) → le#(x, y)

minus#(s(x), s(y)) → minus#(x, y)

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

Problem 2: 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(0, x)0minus(s(x), s(y))minus(x, y)
gcd(0, y)ygcd(s(x), 0)s(x)
gcd(s(x), s(y))if_gcd(le(y, x), s(x), s(y))if_gcd(true, x, y)gcd(minus(x, y), y)
if_gcd(false, x, y)gcd(minus(y, x), 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)

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