TIMEOUT

The TRS could not be proven terminating. The proof attempt took 60103 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (149ms).
 | – Problem 2 remains open; application of the following processors failed [SubtermCriterion (1ms), DependencyGraph (9ms), PolynomialLinearRange4iUR (2255ms), DependencyGraph (8ms), PolynomialLinearRange8NegiUR (timeout), DependencyGraph (9ms), ReductionPairSAT (16444ms), DependencyGraph (30ms), SizeChangePrinciple (158ms), ForwardNarrowing (1ms), BackwardInstantiation (2ms), ForwardInstantiation (4ms), Propagation (2ms)].
 | – Problem 3 was processed with processor SubtermCriterion (0ms).

The following open problems remain:



Open Dependency Pair Problem 2

Dependency Pairs

cond1#(true, x, y) → cond2#(gr(x, 0), x, y)cond3#(false, x, y) → cond1#(or(gr(x, 0), gr(y, 0)), x, y)
cond3#(true, x, y) → cond1#(or(gr(x, 0), gr(y, 0)), x, p(y))cond2#(true, x, y) → cond1#(or(gr(x, 0), gr(y, 0)), p(x), y)
cond2#(false, x, y) → cond3#(gr(y, 0), x, y)

Rewrite Rules

cond1(true, x, y) → cond2(gr(x, 0), x, y)cond2(true, x, y) → cond1(or(gr(x, 0), gr(y, 0)), p(x), y)
cond2(false, x, y) → cond3(gr(y, 0), x, y)cond3(true, x, y) → cond1(or(gr(x, 0), gr(y, 0)), x, p(y))
cond3(false, x, y) → cond1(or(gr(x, 0), gr(y, 0)), x, y)gr(0, x) → false
gr(s(x), 0) → truegr(s(x), s(y)) → gr(x, y)
or(false, false) → falseor(true, x) → true
or(x, true) → truep(0) → 0
p(s(x)) → x

Original Signature

Termination of terms over the following signature is verified: cond2, cond3, 0, s, or, p, false, true, gr, cond1


Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

cond1#(true, x, y) → cond2#(gr(x, 0), x, y)cond3#(true, x, y) → cond1#(or(gr(x, 0), gr(y, 0)), x, p(y))
cond3#(false, x, y) → or#(gr(x, 0), gr(y, 0))cond2#(true, x, y) → gr#(y, 0)
cond3#(true, x, y) → gr#(x, 0)cond3#(false, x, y) → gr#(y, 0)
cond3#(true, x, y) → gr#(y, 0)cond2#(true, x, y) → cond1#(or(gr(x, 0), gr(y, 0)), p(x), y)
cond2#(true, x, y) → or#(gr(x, 0), gr(y, 0))cond2#(true, x, y) → gr#(x, 0)
cond2#(false, x, y) → cond3#(gr(y, 0), x, y)cond3#(true, x, y) → or#(gr(x, 0), gr(y, 0))
cond3#(false, x, y) → cond1#(or(gr(x, 0), gr(y, 0)), x, y)cond2#(false, x, y) → gr#(y, 0)
cond3#(false, x, y) → gr#(x, 0)cond2#(true, x, y) → p#(x)
cond3#(true, x, y) → p#(y)gr#(s(x), s(y)) → gr#(x, y)
cond1#(true, x, y) → gr#(x, 0)

Rewrite Rules

cond1(true, x, y) → cond2(gr(x, 0), x, y)cond2(true, x, y) → cond1(or(gr(x, 0), gr(y, 0)), p(x), y)
cond2(false, x, y) → cond3(gr(y, 0), x, y)cond3(true, x, y) → cond1(or(gr(x, 0), gr(y, 0)), x, p(y))
cond3(false, x, y) → cond1(or(gr(x, 0), gr(y, 0)), x, y)gr(0, x) → false
gr(s(x), 0) → truegr(s(x), s(y)) → gr(x, y)
or(false, false) → falseor(true, x) → true
or(x, true) → truep(0) → 0
p(s(x)) → x

Original Signature

Termination of terms over the following signature is verified: cond2, cond3, 0, s, or, p, true, false, gr, cond1

Strategy


The following SCCs where found

gr#(s(x), s(y)) → gr#(x, y)

cond1#(true, x, y) → cond2#(gr(x, 0), x, y)cond3#(false, x, y) → cond1#(or(gr(x, 0), gr(y, 0)), x, y)
cond3#(true, x, y) → cond1#(or(gr(x, 0), gr(y, 0)), x, p(y))cond2#(true, x, y) → cond1#(or(gr(x, 0), gr(y, 0)), p(x), y)
cond2#(false, x, y) → cond3#(gr(y, 0), x, y)

Problem 3: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

gr#(s(x), s(y)) → gr#(x, y)

Rewrite Rules

cond1(true, x, y) → cond2(gr(x, 0), x, y)cond2(true, x, y) → cond1(or(gr(x, 0), gr(y, 0)), p(x), y)
cond2(false, x, y) → cond3(gr(y, 0), x, y)cond3(true, x, y) → cond1(or(gr(x, 0), gr(y, 0)), x, p(y))
cond3(false, x, y) → cond1(or(gr(x, 0), gr(y, 0)), x, y)gr(0, x) → false
gr(s(x), 0) → truegr(s(x), s(y)) → gr(x, y)
or(false, false) → falseor(true, x) → true
or(x, true) → truep(0) → 0
p(s(x)) → x

Original Signature

Termination of terms over the following signature is verified: cond2, cond3, 0, s, or, p, true, false, gr, cond1

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

gr#(s(x), s(y)) → gr#(x, y)