TIMEOUT

The TRS could not be proven terminating. The proof attempt took 60131 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (651ms).
 | – Problem 2 was processed with processor SubtermCriterion (2ms).
 | – Problem 3 was processed with processor SubtermCriterion (4ms).
 | – Problem 4 was processed with processor SubtermCriterion (1ms).
 | – Problem 5 was processed with processor SubtermCriterion (0ms).
 | – Problem 6 was processed with processor PolynomialLinearRange4iUR (127ms).
 | – Problem 7 was processed with processor PolynomialLinearRange4iUR (2550ms).
 | – Problem 8 was processed with processor SubtermCriterion (1ms).
 | – Problem 9 remains open; application of the following processors failed [SubtermCriterion (0ms), DependencyGraph (6ms), PolynomialLinearRange4iUR (1400ms), DependencyGraph (6ms), PolynomialLinearRange4iUR (1088ms), DependencyGraph (5ms), PolynomialLinearRange8NegiUR (3765ms), DependencyGraph (47ms), ReductionPairSAT (1598ms), DependencyGraph (5ms), SizeChangePrinciple (476ms), ForwardNarrowing (1ms), BackwardInstantiation (4ms), ForwardInstantiation (2ms), Propagation (2ms)].

The following open problems remain:



Open Dependency Pair Problem 9

Dependency Pairs

p#(s(x), s(y))p#(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))p#(id(x), s(y))p#(x, if(gt(s(y), y), y, s(y)))
p#(s(x), x)p#(if(gt(x, x), id(x), id(x)), s(x))

Rewrite Rules

g(s(x), s(y))if(and(f(s(x)), f(s(y))), t(g(k(minus(m(x, y), n(x, y)), s(s(0))), k(n(s(x), s(y)), s(s(0))))), g(minus(m(x, y), n(x, y)), n(s(x), s(y))))n(0, y)0
n(x, 0)0n(s(x), s(y))s(n(x, y))
m(0, y)ym(x, 0)x
m(s(x), s(y))s(m(x, y))k(0, s(y))0
k(s(x), s(y))s(k(minus(x, y), s(y)))t(x)p(x, x)
p(s(x), s(y))s(s(p(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))p(s(x), x)p(if(gt(x, x), id(x), id(x)), s(x))
p(0, y)yp(id(x), s(y))s(p(x, if(gt(s(y), y), y, s(y))))
minus(x, 0)xminus(s(x), s(y))minus(x, y)
id(x)xif(true, x, y)x
if(false, x, y)ynot(x)if(x, false, true)
and(x, false)falseand(true, true)true
f(0)truef(s(x))h(x)
h(0)falseh(s(x))f(x)
gt(s(x), 0)truegt(0, y)false
gt(s(x), s(y))gt(x, y)

Original Signature

Termination of terms over the following signature is verified: f, g, minus, n, true, m, k, h, and, not, id, 0, t, s, if, p, false, gt


Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

g#(s(x), s(y))m#(x, y)g#(s(x), s(y))g#(minus(m(x, y), n(x, y)), n(s(x), s(y)))
p#(id(x), s(y))if#(gt(s(y), y), y, s(y))p#(id(x), s(y))p#(x, if(gt(s(y), y), y, s(y)))
n#(s(x), s(y))n#(x, y)g#(s(x), s(y))n#(x, y)
m#(s(x), s(y))m#(x, y)p#(s(x), x)if#(gt(x, x), id(x), id(x))
g#(s(x), s(y))t#(g(k(minus(m(x, y), n(x, y)), s(s(0))), k(n(s(x), s(y)), s(s(0)))))t#(x)p#(x, x)
g#(s(x), s(y))if#(and(f(s(x)), f(s(y))), t(g(k(minus(m(x, y), n(x, y)), s(s(0))), k(n(s(x), s(y)), s(s(0))))), g(minus(m(x, y), n(x, y)), n(s(x), s(y))))f#(s(x))h#(x)
g#(s(x), s(y))f#(s(x))p#(s(x), s(y))if#(gt(x, y), x, y)
g#(s(x), s(y))k#(n(s(x), s(y)), s(s(0)))gt#(s(x), s(y))gt#(x, y)
g#(s(x), s(y))minus#(m(x, y), n(x, y))g#(s(x), s(y))f#(s(y))
p#(s(x), x)id#(x)h#(s(x))f#(x)
not#(x)if#(x, false, true)p#(s(x), x)p#(if(gt(x, x), id(x), id(x)), s(x))
p#(s(x), s(y))p#(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))p#(s(x), s(y))not#(gt(x, y))
p#(s(x), x)gt#(x, x)p#(s(x), s(y))id#(x)
p#(s(x), s(y))id#(y)p#(s(x), s(y))gt#(x, y)
p#(id(x), s(y))gt#(s(y), y)g#(s(x), s(y))and#(f(s(x)), f(s(y)))
k#(s(x), s(y))minus#(x, y)minus#(s(x), s(y))minus#(x, y)
g#(s(x), s(y))k#(minus(m(x, y), n(x, y)), s(s(0)))p#(s(x), s(y))if#(not(gt(x, y)), id(x), id(y))
k#(s(x), s(y))k#(minus(x, y), s(y))g#(s(x), s(y))g#(k(minus(m(x, y), n(x, y)), s(s(0))), k(n(s(x), s(y)), s(s(0))))
g#(s(x), s(y))n#(s(x), s(y))

Rewrite Rules

g(s(x), s(y))if(and(f(s(x)), f(s(y))), t(g(k(minus(m(x, y), n(x, y)), s(s(0))), k(n(s(x), s(y)), s(s(0))))), g(minus(m(x, y), n(x, y)), n(s(x), s(y))))n(0, y)0
n(x, 0)0n(s(x), s(y))s(n(x, y))
m(0, y)ym(x, 0)x
m(s(x), s(y))s(m(x, y))k(0, s(y))0
k(s(x), s(y))s(k(minus(x, y), s(y)))t(x)p(x, x)
p(s(x), s(y))s(s(p(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))p(s(x), x)p(if(gt(x, x), id(x), id(x)), s(x))
p(0, y)yp(id(x), s(y))s(p(x, if(gt(s(y), y), y, s(y))))
minus(x, 0)xminus(s(x), s(y))minus(x, y)
id(x)xif(true, x, y)x
if(false, x, y)ynot(x)if(x, false, true)
and(x, false)falseand(true, true)true
f(0)truef(s(x))h(x)
h(0)falseh(s(x))f(x)
gt(s(x), 0)truegt(0, y)false
gt(s(x), s(y))gt(x, y)

Original Signature

Termination of terms over the following signature is verified: f, g, minus, n, true, m, k, h, and, not, id, 0, t, s, if, p, false, gt

Strategy


The following SCCs where found

n#(s(x), s(y)) → n#(x, y)

gt#(s(x), s(y)) → gt#(x, y)

minus#(s(x), s(y)) → minus#(x, y)

m#(s(x), s(y)) → m#(x, y)

p#(s(x), s(y)) → p#(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))p#(id(x), s(y)) → p#(x, if(gt(s(y), y), y, s(y)))
p#(s(x), x) → p#(if(gt(x, x), id(x), id(x)), s(x))

k#(s(x), s(y)) → k#(minus(x, y), s(y))

h#(s(x)) → f#(x)f#(s(x)) → h#(x)

g#(s(x), s(y)) → g#(minus(m(x, y), n(x, y)), n(s(x), s(y)))g#(s(x), s(y)) → g#(k(minus(m(x, y), n(x, y)), s(s(0))), k(n(s(x), s(y)), s(s(0))))

Problem 2: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

gt#(s(x), s(y))gt#(x, y)

Rewrite Rules

g(s(x), s(y))if(and(f(s(x)), f(s(y))), t(g(k(minus(m(x, y), n(x, y)), s(s(0))), k(n(s(x), s(y)), s(s(0))))), g(minus(m(x, y), n(x, y)), n(s(x), s(y))))n(0, y)0
n(x, 0)0n(s(x), s(y))s(n(x, y))
m(0, y)ym(x, 0)x
m(s(x), s(y))s(m(x, y))k(0, s(y))0
k(s(x), s(y))s(k(minus(x, y), s(y)))t(x)p(x, x)
p(s(x), s(y))s(s(p(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))p(s(x), x)p(if(gt(x, x), id(x), id(x)), s(x))
p(0, y)yp(id(x), s(y))s(p(x, if(gt(s(y), y), y, s(y))))
minus(x, 0)xminus(s(x), s(y))minus(x, y)
id(x)xif(true, x, y)x
if(false, x, y)ynot(x)if(x, false, true)
and(x, false)falseand(true, true)true
f(0)truef(s(x))h(x)
h(0)falseh(s(x))f(x)
gt(s(x), 0)truegt(0, y)false
gt(s(x), s(y))gt(x, y)

Original Signature

Termination of terms over the following signature is verified: f, g, minus, n, true, m, k, h, and, not, id, 0, t, s, if, p, false, gt

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

gt#(s(x), s(y))gt#(x, y)

Problem 3: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

n#(s(x), s(y))n#(x, y)

Rewrite Rules

g(s(x), s(y))if(and(f(s(x)), f(s(y))), t(g(k(minus(m(x, y), n(x, y)), s(s(0))), k(n(s(x), s(y)), s(s(0))))), g(minus(m(x, y), n(x, y)), n(s(x), s(y))))n(0, y)0
n(x, 0)0n(s(x), s(y))s(n(x, y))
m(0, y)ym(x, 0)x
m(s(x), s(y))s(m(x, y))k(0, s(y))0
k(s(x), s(y))s(k(minus(x, y), s(y)))t(x)p(x, x)
p(s(x), s(y))s(s(p(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))p(s(x), x)p(if(gt(x, x), id(x), id(x)), s(x))
p(0, y)yp(id(x), s(y))s(p(x, if(gt(s(y), y), y, s(y))))
minus(x, 0)xminus(s(x), s(y))minus(x, y)
id(x)xif(true, x, y)x
if(false, x, y)ynot(x)if(x, false, true)
and(x, false)falseand(true, true)true
f(0)truef(s(x))h(x)
h(0)falseh(s(x))f(x)
gt(s(x), 0)truegt(0, y)false
gt(s(x), s(y))gt(x, y)

Original Signature

Termination of terms over the following signature is verified: f, g, minus, n, true, m, k, h, and, not, id, 0, t, s, if, p, false, gt

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

n#(s(x), s(y))n#(x, y)

Problem 4: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

h#(s(x))f#(x)f#(s(x))h#(x)

Rewrite Rules

g(s(x), s(y))if(and(f(s(x)), f(s(y))), t(g(k(minus(m(x, y), n(x, y)), s(s(0))), k(n(s(x), s(y)), s(s(0))))), g(minus(m(x, y), n(x, y)), n(s(x), s(y))))n(0, y)0
n(x, 0)0n(s(x), s(y))s(n(x, y))
m(0, y)ym(x, 0)x
m(s(x), s(y))s(m(x, y))k(0, s(y))0
k(s(x), s(y))s(k(minus(x, y), s(y)))t(x)p(x, x)
p(s(x), s(y))s(s(p(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))p(s(x), x)p(if(gt(x, x), id(x), id(x)), s(x))
p(0, y)yp(id(x), s(y))s(p(x, if(gt(s(y), y), y, s(y))))
minus(x, 0)xminus(s(x), s(y))minus(x, y)
id(x)xif(true, x, y)x
if(false, x, y)ynot(x)if(x, false, true)
and(x, false)falseand(true, true)true
f(0)truef(s(x))h(x)
h(0)falseh(s(x))f(x)
gt(s(x), 0)truegt(0, y)false
gt(s(x), s(y))gt(x, y)

Original Signature

Termination of terms over the following signature is verified: f, g, minus, n, true, m, k, h, and, not, id, 0, t, s, if, p, false, gt

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

h#(s(x))f#(x)f#(s(x))h#(x)

Problem 5: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

m#(s(x), s(y))m#(x, y)

Rewrite Rules

g(s(x), s(y))if(and(f(s(x)), f(s(y))), t(g(k(minus(m(x, y), n(x, y)), s(s(0))), k(n(s(x), s(y)), s(s(0))))), g(minus(m(x, y), n(x, y)), n(s(x), s(y))))n(0, y)0
n(x, 0)0n(s(x), s(y))s(n(x, y))
m(0, y)ym(x, 0)x
m(s(x), s(y))s(m(x, y))k(0, s(y))0
k(s(x), s(y))s(k(minus(x, y), s(y)))t(x)p(x, x)
p(s(x), s(y))s(s(p(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))p(s(x), x)p(if(gt(x, x), id(x), id(x)), s(x))
p(0, y)yp(id(x), s(y))s(p(x, if(gt(s(y), y), y, s(y))))
minus(x, 0)xminus(s(x), s(y))minus(x, y)
id(x)xif(true, x, y)x
if(false, x, y)ynot(x)if(x, false, true)
and(x, false)falseand(true, true)true
f(0)truef(s(x))h(x)
h(0)falseh(s(x))f(x)
gt(s(x), 0)truegt(0, y)false
gt(s(x), s(y))gt(x, y)

Original Signature

Termination of terms over the following signature is verified: f, g, minus, n, true, m, k, h, and, not, id, 0, t, s, if, p, false, gt

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

m#(s(x), s(y))m#(x, y)

Problem 6: PolynomialLinearRange4iUR



Dependency Pair Problem

Dependency Pairs

k#(s(x), s(y))k#(minus(x, y), s(y))

Rewrite Rules

g(s(x), s(y))if(and(f(s(x)), f(s(y))), t(g(k(minus(m(x, y), n(x, y)), s(s(0))), k(n(s(x), s(y)), s(s(0))))), g(minus(m(x, y), n(x, y)), n(s(x), s(y))))n(0, y)0
n(x, 0)0n(s(x), s(y))s(n(x, y))
m(0, y)ym(x, 0)x
m(s(x), s(y))s(m(x, y))k(0, s(y))0
k(s(x), s(y))s(k(minus(x, y), s(y)))t(x)p(x, x)
p(s(x), s(y))s(s(p(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))p(s(x), x)p(if(gt(x, x), id(x), id(x)), s(x))
p(0, y)yp(id(x), s(y))s(p(x, if(gt(s(y), y), y, s(y))))
minus(x, 0)xminus(s(x), s(y))minus(x, y)
id(x)xif(true, x, y)x
if(false, x, y)ynot(x)if(x, false, true)
and(x, false)falseand(true, true)true
f(0)truef(s(x))h(x)
h(0)falseh(s(x))f(x)
gt(s(x), 0)truegt(0, y)false
gt(s(x), s(y))gt(x, y)

Original Signature

Termination of terms over the following signature is verified: f, g, minus, n, true, m, k, h, and, not, id, 0, t, s, if, p, false, gt

Strategy


Polynomial Interpretation

Improved Usable rules

minus(s(x), s(y))minus(x, y)minus(x, 0)x

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

k#(s(x), s(y))k#(minus(x, y), s(y))

Problem 7: PolynomialLinearRange4iUR



Dependency Pair Problem

Dependency Pairs

g#(s(x), s(y))g#(minus(m(x, y), n(x, y)), n(s(x), s(y)))g#(s(x), s(y))g#(k(minus(m(x, y), n(x, y)), s(s(0))), k(n(s(x), s(y)), s(s(0))))

Rewrite Rules

g(s(x), s(y))if(and(f(s(x)), f(s(y))), t(g(k(minus(m(x, y), n(x, y)), s(s(0))), k(n(s(x), s(y)), s(s(0))))), g(minus(m(x, y), n(x, y)), n(s(x), s(y))))n(0, y)0
n(x, 0)0n(s(x), s(y))s(n(x, y))
m(0, y)ym(x, 0)x
m(s(x), s(y))s(m(x, y))k(0, s(y))0
k(s(x), s(y))s(k(minus(x, y), s(y)))t(x)p(x, x)
p(s(x), s(y))s(s(p(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))p(s(x), x)p(if(gt(x, x), id(x), id(x)), s(x))
p(0, y)yp(id(x), s(y))s(p(x, if(gt(s(y), y), y, s(y))))
minus(x, 0)xminus(s(x), s(y))minus(x, y)
id(x)xif(true, x, y)x
if(false, x, y)ynot(x)if(x, false, true)
and(x, false)falseand(true, true)true
f(0)truef(s(x))h(x)
h(0)falseh(s(x))f(x)
gt(s(x), 0)truegt(0, y)false
gt(s(x), s(y))gt(x, y)

Original Signature

Termination of terms over the following signature is verified: f, g, minus, n, true, m, k, h, and, not, id, 0, t, s, if, p, false, gt

Strategy


Polynomial Interpretation

Improved Usable rules

minus(s(x), s(y))minus(x, y)m(x, 0)x
m(s(x), s(y))s(m(x, y))m(0, y)y
k(0, s(y))0minus(x, 0)x
n(s(x), s(y))s(n(x, y))k(s(x), s(y))s(k(minus(x, y), s(y)))
n(x, 0)0n(0, y)0

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

g#(s(x), s(y))g#(minus(m(x, y), n(x, y)), n(s(x), s(y)))g#(s(x), s(y))g#(k(minus(m(x, y), n(x, y)), s(s(0))), k(n(s(x), s(y)), s(s(0))))

Problem 8: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

minus#(s(x), s(y))minus#(x, y)

Rewrite Rules

g(s(x), s(y))if(and(f(s(x)), f(s(y))), t(g(k(minus(m(x, y), n(x, y)), s(s(0))), k(n(s(x), s(y)), s(s(0))))), g(minus(m(x, y), n(x, y)), n(s(x), s(y))))n(0, y)0
n(x, 0)0n(s(x), s(y))s(n(x, y))
m(0, y)ym(x, 0)x
m(s(x), s(y))s(m(x, y))k(0, s(y))0
k(s(x), s(y))s(k(minus(x, y), s(y)))t(x)p(x, x)
p(s(x), s(y))s(s(p(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y)))))p(s(x), x)p(if(gt(x, x), id(x), id(x)), s(x))
p(0, y)yp(id(x), s(y))s(p(x, if(gt(s(y), y), y, s(y))))
minus(x, 0)xminus(s(x), s(y))minus(x, y)
id(x)xif(true, x, y)x
if(false, x, y)ynot(x)if(x, false, true)
and(x, false)falseand(true, true)true
f(0)truef(s(x))h(x)
h(0)falseh(s(x))f(x)
gt(s(x), 0)truegt(0, y)false
gt(s(x), s(y))gt(x, y)

Original Signature

Termination of terms over the following signature is verified: f, g, minus, n, true, m, k, h, and, not, id, 0, t, s, if, p, false, gt

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

minus#(s(x), s(y))minus#(x, y)