TIMEOUT

The TRS could not be proven terminating. The proof attempt took 60001 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (133ms).
 | – Problem 2 was processed with processor SubtermCriterion (1ms).
 | – Problem 3 was processed with processor SubtermCriterion (1ms).
 | – Problem 4 was processed with processor SubtermCriterion (0ms).
 | – Problem 5 remains open; application of the following processors failed [SubtermCriterion (2ms), DependencyGraph (10ms), PolynomialLinearRange4iUR (1541ms), DependencyGraph (9ms), PolynomialLinearRange8NegiUR (29997ms), DependencyGraph (9ms), ReductionPairSAT (5875ms), DependencyGraph (47ms), SizeChangePrinciple (805ms), ForwardNarrowing (3ms), BackwardInstantiation (2ms), ForwardInstantiation (4ms), Propagation (1ms)].

The following open problems remain:



Open Dependency Pair Problem 5

Dependency Pairs

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

Rewrite Rules

cond1(true, x, y) → cond2(gr(x, y), x, y)cond2(true, x, y) → cond1(gr(add(x, y), 0), p(x), y)
cond2(false, x, y) → cond3(eq(x, y), x, y)cond3(true, x, y) → cond1(gr(add(x, y), 0), p(x), y)
cond3(false, x, y) → cond1(gr(add(x, y), 0), x, p(y))gr(0, x) → false
gr(s(x), 0) → truegr(s(x), s(y)) → gr(x, y)
add(0, x) → xadd(s(x), y) → s(add(x, y))
eq(0, 0) → trueeq(0, s(x)) → false
eq(s(x), 0) → falseeq(s(x), s(y)) → eq(x, y)
p(0) → 0p(s(x)) → x

Original Signature

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


Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

cond1(true, x, y) → cond2(gr(x, y), x, y)cond2(true, x, y) → cond1(gr(add(x, y), 0), p(x), y)
cond2(false, x, y) → cond3(eq(x, y), x, y)cond3(true, x, y) → cond1(gr(add(x, y), 0), p(x), y)
cond3(false, x, y) → cond1(gr(add(x, y), 0), x, p(y))gr(0, x) → false
gr(s(x), 0) → truegr(s(x), s(y)) → gr(x, y)
add(0, x) → xadd(s(x), y) → s(add(x, y))
eq(0, 0) → trueeq(0, s(x)) → false
eq(s(x), 0) → falseeq(s(x), s(y)) → eq(x, y)
p(0) → 0p(s(x)) → x

Original Signature

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

Strategy


The following SCCs where found

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

gr#(s(x), s(y)) → gr#(x, y)

add#(s(x), y) → add#(x, y)

eq#(s(x), s(y)) → eq#(x, y)

Problem 2: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

gr#(s(x), s(y)) → gr#(x, y)

Rewrite Rules

cond1(true, x, y) → cond2(gr(x, y), x, y)cond2(true, x, y) → cond1(gr(add(x, y), 0), p(x), y)
cond2(false, x, y) → cond3(eq(x, y), x, y)cond3(true, x, y) → cond1(gr(add(x, y), 0), p(x), y)
cond3(false, x, y) → cond1(gr(add(x, y), 0), x, p(y))gr(0, x) → false
gr(s(x), 0) → truegr(s(x), s(y)) → gr(x, y)
add(0, x) → xadd(s(x), y) → s(add(x, y))
eq(0, 0) → trueeq(0, s(x)) → false
eq(s(x), 0) → falseeq(s(x), s(y)) → eq(x, y)
p(0) → 0p(s(x)) → x

Original Signature

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

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

gr#(s(x), s(y)) → gr#(x, y)

Problem 3: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

add#(s(x), y) → add#(x, y)

Rewrite Rules

cond1(true, x, y) → cond2(gr(x, y), x, y)cond2(true, x, y) → cond1(gr(add(x, y), 0), p(x), y)
cond2(false, x, y) → cond3(eq(x, y), x, y)cond3(true, x, y) → cond1(gr(add(x, y), 0), p(x), y)
cond3(false, x, y) → cond1(gr(add(x, y), 0), x, p(y))gr(0, x) → false
gr(s(x), 0) → truegr(s(x), s(y)) → gr(x, y)
add(0, x) → xadd(s(x), y) → s(add(x, y))
eq(0, 0) → trueeq(0, s(x)) → false
eq(s(x), 0) → falseeq(s(x), s(y)) → eq(x, y)
p(0) → 0p(s(x)) → x

Original Signature

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

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

add#(s(x), y) → add#(x, y)

Problem 4: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

eq#(s(x), s(y)) → eq#(x, y)

Rewrite Rules

cond1(true, x, y) → cond2(gr(x, y), x, y)cond2(true, x, y) → cond1(gr(add(x, y), 0), p(x), y)
cond2(false, x, y) → cond3(eq(x, y), x, y)cond3(true, x, y) → cond1(gr(add(x, y), 0), p(x), y)
cond3(false, x, y) → cond1(gr(add(x, y), 0), x, p(y))gr(0, x) → false
gr(s(x), 0) → truegr(s(x), s(y)) → gr(x, y)
add(0, x) → xadd(s(x), y) → s(add(x, y))
eq(0, 0) → trueeq(0, s(x)) → false
eq(s(x), 0) → falseeq(s(x), s(y)) → eq(x, y)
p(0) → 0p(s(x)) → x

Original Signature

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

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

eq#(s(x), s(y)) → eq#(x, y)