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)