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)