TIMEOUT
The TRS could not be proven terminating. The proof attempt took 60000 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (393ms).
| – Problem 2 remains open; application of the following processors failed [SubtermCriterion (1ms), DependencyGraph (60ms), PolynomialLinearRange4iUR (5000ms), DependencyGraph (41ms), PolynomialLinearRange8NegiUR (15000ms), DependencyGraph (40ms), ReductionPairSAT (3627ms), DependencyGraph (40ms), ReductionPairSAT (3477ms), DependencyGraph (42ms), SizeChangePrinciple (timeout)].
| – Problem 3 was processed with processor SubtermCriterion (0ms).
| – Problem 4 was processed with processor ReductionPairSAT (218ms).
| | – Problem 6 remains open; application of the following processors failed [DependencyGraph (2ms), ReductionPairSAT (98ms), DependencyGraph (2ms)].
| – Problem 5 was processed with processor SubtermCriterion (1ms).
The following open problems remain:
Open Dependency Pair Problem 2
Dependency Pairs
| divides#(y, x) | → | div#(x, y) | | if#(false, x, y) | → | pr#(x, y) |
| div#(div(x, y), z) | → | zero#(y) | | pr#(x, s(s(y))) | → | divides#(s(s(y)), x) |
| zero#(s(x)) | → | if#(eq(x, s(0)), plus(zero(0), 0), s(plus(0, zero(0)))) | | quot#(x, 0, s(z)) | → | div#(x, s(z)) |
| div#(x, y) | → | quot#(x, y, y) | | pr#(x, s(s(y))) | → | if#(divides(s(s(y)), x), x, s(y)) |
| quot#(s(x), s(y), z) | → | quot#(x, y, z) | | div#(div(x, y), z) | → | div#(x, times(zero(y), z)) |
Rewrite Rules
| p(0) | → | 0 | | p(s(x)) | → | x |
| plus(x, 0) | → | x | | plus(0, y) | → | y |
| plus(s(x), y) | → | s(plus(x, y)) | | plus(s(x), y) | → | s(plus(p(s(x)), y)) |
| plus(x, s(y)) | → | s(plus(x, p(s(y)))) | | times(0, y) | → | 0 |
| times(s(0), y) | → | y | | times(s(x), y) | → | plus(y, times(x, y)) |
| div(0, y) | → | 0 | | div(x, y) | → | quot(x, y, y) |
| quot(zero(y), s(y), z) | → | 0 | | quot(s(x), s(y), z) | → | quot(x, y, z) |
| quot(x, 0, s(z)) | → | s(div(x, s(z))) | | div(div(x, y), z) | → | div(x, times(zero(y), z)) |
| eq(0, 0) | → | true | | eq(s(x), 0) | → | false |
| eq(0, s(y)) | → | false | | eq(s(x), s(y)) | → | eq(x, y) |
| divides(y, x) | → | eq(x, times(div(x, y), y)) | | prime(s(s(x))) | → | pr(s(s(x)), s(x)) |
| pr(x, s(0)) | → | true | | pr(x, s(s(y))) | → | if(divides(s(s(y)), x), x, s(y)) |
| if(true, x, y) | → | false | | if(false, x, y) | → | pr(x, y) |
| zero(div(x, x)) | → | x | | zero(divides(x, x)) | → | x |
| zero(times(x, x)) | → | x | | zero(quot(x, x, x)) | → | x |
| zero(s(x)) | → | if(eq(x, s(0)), plus(zero(0), 0), s(plus(0, zero(0)))) |
Original Signature
Termination of terms over the following signature is verified: plus, div, true, divides, zero, prime, 0, s, times, if, p, false, quot, pr, eq
Open Dependency Pair Problem 6
Dependency Pairs
| plus#(s(x), y) | → | plus#(p(s(x)), y) | | plus#(x, s(y)) | → | plus#(x, p(s(y))) |
Rewrite Rules
| p(0) | → | 0 | | p(s(x)) | → | x |
| plus(x, 0) | → | x | | plus(0, y) | → | y |
| plus(s(x), y) | → | s(plus(x, y)) | | plus(s(x), y) | → | s(plus(p(s(x)), y)) |
| plus(x, s(y)) | → | s(plus(x, p(s(y)))) | | times(0, y) | → | 0 |
| times(s(0), y) | → | y | | times(s(x), y) | → | plus(y, times(x, y)) |
| div(0, y) | → | 0 | | div(x, y) | → | quot(x, y, y) |
| quot(zero(y), s(y), z) | → | 0 | | quot(s(x), s(y), z) | → | quot(x, y, z) |
| quot(x, 0, s(z)) | → | s(div(x, s(z))) | | div(div(x, y), z) | → | div(x, times(zero(y), z)) |
| eq(0, 0) | → | true | | eq(s(x), 0) | → | false |
| eq(0, s(y)) | → | false | | eq(s(x), s(y)) | → | eq(x, y) |
| divides(y, x) | → | eq(x, times(div(x, y), y)) | | prime(s(s(x))) | → | pr(s(s(x)), s(x)) |
| pr(x, s(0)) | → | true | | pr(x, s(s(y))) | → | if(divides(s(s(y)), x), x, s(y)) |
| if(true, x, y) | → | false | | if(false, x, y) | → | pr(x, y) |
| zero(div(x, x)) | → | x | | zero(divides(x, x)) | → | x |
| zero(times(x, x)) | → | x | | zero(quot(x, x, x)) | → | x |
| zero(s(x)) | → | if(eq(x, s(0)), plus(zero(0), 0), s(plus(0, zero(0)))) |
Original Signature
Termination of terms over the following signature is verified: plus, div, true, divides, zero, prime, 0, s, times, if, p, false, quot, pr, eq
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| zero#(s(x)) | → | plus#(0, zero(0)) | | pr#(x, s(s(y))) | → | divides#(s(s(y)), x) |
| zero#(s(x)) | → | if#(eq(x, s(0)), plus(zero(0), 0), s(plus(0, zero(0)))) | | pr#(x, s(s(y))) | → | if#(divides(s(s(y)), x), x, s(y)) |
| divides#(y, x) | → | eq#(x, times(div(x, y), y)) | | if#(false, x, y) | → | pr#(x, y) |
| plus#(s(x), y) | → | plus#(x, y) | | plus#(s(x), y) | → | plus#(p(s(x)), y) |
| prime#(s(s(x))) | → | pr#(s(s(x)), s(x)) | | eq#(s(x), s(y)) | → | eq#(x, y) |
| plus#(x, s(y)) | → | p#(s(y)) | | divides#(y, x) | → | div#(x, y) |
| div#(div(x, y), z) | → | zero#(y) | | zero#(s(x)) | → | plus#(zero(0), 0) |
| times#(s(x), y) | → | times#(x, y) | | zero#(s(x)) | → | eq#(x, s(0)) |
| div#(div(x, y), z) | → | times#(zero(y), z) | | div#(div(x, y), z) | → | div#(x, times(zero(y), z)) |
| quot#(s(x), s(y), z) | → | quot#(x, y, z) | | times#(s(x), y) | → | plus#(y, times(x, y)) |
| quot#(x, 0, s(z)) | → | div#(x, s(z)) | | divides#(y, x) | → | times#(div(x, y), y) |
| div#(x, y) | → | quot#(x, y, y) | | zero#(s(x)) | → | zero#(0) |
| plus#(s(x), y) | → | p#(s(x)) | | plus#(x, s(y)) | → | plus#(x, p(s(y))) |
Rewrite Rules
| p(0) | → | 0 | | p(s(x)) | → | x |
| plus(x, 0) | → | x | | plus(0, y) | → | y |
| plus(s(x), y) | → | s(plus(x, y)) | | plus(s(x), y) | → | s(plus(p(s(x)), y)) |
| plus(x, s(y)) | → | s(plus(x, p(s(y)))) | | times(0, y) | → | 0 |
| times(s(0), y) | → | y | | times(s(x), y) | → | plus(y, times(x, y)) |
| div(0, y) | → | 0 | | div(x, y) | → | quot(x, y, y) |
| quot(zero(y), s(y), z) | → | 0 | | quot(s(x), s(y), z) | → | quot(x, y, z) |
| quot(x, 0, s(z)) | → | s(div(x, s(z))) | | div(div(x, y), z) | → | div(x, times(zero(y), z)) |
| eq(0, 0) | → | true | | eq(s(x), 0) | → | false |
| eq(0, s(y)) | → | false | | eq(s(x), s(y)) | → | eq(x, y) |
| divides(y, x) | → | eq(x, times(div(x, y), y)) | | prime(s(s(x))) | → | pr(s(s(x)), s(x)) |
| pr(x, s(0)) | → | true | | pr(x, s(s(y))) | → | if(divides(s(s(y)), x), x, s(y)) |
| if(true, x, y) | → | false | | if(false, x, y) | → | pr(x, y) |
| zero(div(x, x)) | → | x | | zero(divides(x, x)) | → | x |
| zero(times(x, x)) | → | x | | zero(quot(x, x, x)) | → | x |
| zero(s(x)) | → | if(eq(x, s(0)), plus(zero(0), 0), s(plus(0, zero(0)))) |
Original Signature
Termination of terms over the following signature is verified: plus, div, true, divides, zero, prime, 0, s, times, if, p, false, quot, pr, eq
Strategy
The following SCCs where found
| times#(s(x), y) → times#(x, y) |
| plus#(s(x), y) → plus#(x, y) | plus#(s(x), y) → plus#(p(s(x)), y) |
| plus#(x, s(y)) → plus#(x, p(s(y))) |
| eq#(s(x), s(y)) → eq#(x, y) |
| divides#(y, x) → div#(x, y) | div#(div(x, y), z) → zero#(y) |
| if#(false, x, y) → pr#(x, y) | pr#(x, s(s(y))) → divides#(s(s(y)), x) |
| zero#(s(x)) → if#(eq(x, s(0)), plus(zero(0), 0), s(plus(0, zero(0)))) | quot#(x, 0, s(z)) → div#(x, s(z)) |
| pr#(x, s(s(y))) → if#(divides(s(s(y)), x), x, s(y)) | div#(x, y) → quot#(x, y, y) |
| div#(div(x, y), z) → div#(x, times(zero(y), z)) | quot#(s(x), s(y), z) → quot#(x, y, z) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| times#(s(x), y) | → | times#(x, y) |
Rewrite Rules
| p(0) | → | 0 | | p(s(x)) | → | x |
| plus(x, 0) | → | x | | plus(0, y) | → | y |
| plus(s(x), y) | → | s(plus(x, y)) | | plus(s(x), y) | → | s(plus(p(s(x)), y)) |
| plus(x, s(y)) | → | s(plus(x, p(s(y)))) | | times(0, y) | → | 0 |
| times(s(0), y) | → | y | | times(s(x), y) | → | plus(y, times(x, y)) |
| div(0, y) | → | 0 | | div(x, y) | → | quot(x, y, y) |
| quot(zero(y), s(y), z) | → | 0 | | quot(s(x), s(y), z) | → | quot(x, y, z) |
| quot(x, 0, s(z)) | → | s(div(x, s(z))) | | div(div(x, y), z) | → | div(x, times(zero(y), z)) |
| eq(0, 0) | → | true | | eq(s(x), 0) | → | false |
| eq(0, s(y)) | → | false | | eq(s(x), s(y)) | → | eq(x, y) |
| divides(y, x) | → | eq(x, times(div(x, y), y)) | | prime(s(s(x))) | → | pr(s(s(x)), s(x)) |
| pr(x, s(0)) | → | true | | pr(x, s(s(y))) | → | if(divides(s(s(y)), x), x, s(y)) |
| if(true, x, y) | → | false | | if(false, x, y) | → | pr(x, y) |
| zero(div(x, x)) | → | x | | zero(divides(x, x)) | → | x |
| zero(times(x, x)) | → | x | | zero(quot(x, x, x)) | → | x |
| zero(s(x)) | → | if(eq(x, s(0)), plus(zero(0), 0), s(plus(0, zero(0)))) |
Original Signature
Termination of terms over the following signature is verified: plus, div, true, divides, zero, prime, 0, s, times, if, p, false, quot, pr, eq
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| times#(s(x), y) | → | times#(x, y) |
Problem 4: ReductionPairSAT
Dependency Pair Problem
Dependency Pairs
| plus#(s(x), y) | → | plus#(x, y) | | plus#(s(x), y) | → | plus#(p(s(x)), y) |
| plus#(x, s(y)) | → | plus#(x, p(s(y))) |
Rewrite Rules
| p(0) | → | 0 | | p(s(x)) | → | x |
| plus(x, 0) | → | x | | plus(0, y) | → | y |
| plus(s(x), y) | → | s(plus(x, y)) | | plus(s(x), y) | → | s(plus(p(s(x)), y)) |
| plus(x, s(y)) | → | s(plus(x, p(s(y)))) | | times(0, y) | → | 0 |
| times(s(0), y) | → | y | | times(s(x), y) | → | plus(y, times(x, y)) |
| div(0, y) | → | 0 | | div(x, y) | → | quot(x, y, y) |
| quot(zero(y), s(y), z) | → | 0 | | quot(s(x), s(y), z) | → | quot(x, y, z) |
| quot(x, 0, s(z)) | → | s(div(x, s(z))) | | div(div(x, y), z) | → | div(x, times(zero(y), z)) |
| eq(0, 0) | → | true | | eq(s(x), 0) | → | false |
| eq(0, s(y)) | → | false | | eq(s(x), s(y)) | → | eq(x, y) |
| divides(y, x) | → | eq(x, times(div(x, y), y)) | | prime(s(s(x))) | → | pr(s(s(x)), s(x)) |
| pr(x, s(0)) | → | true | | pr(x, s(s(y))) | → | if(divides(s(s(y)), x), x, s(y)) |
| if(true, x, y) | → | false | | if(false, x, y) | → | pr(x, y) |
| zero(div(x, x)) | → | x | | zero(divides(x, x)) | → | x |
| zero(times(x, x)) | → | x | | zero(quot(x, x, x)) | → | x |
| zero(s(x)) | → | if(eq(x, s(0)), plus(zero(0), 0), s(plus(0, zero(0)))) |
Original Signature
Termination of terms over the following signature is verified: plus, div, true, divides, zero, prime, 0, s, times, if, p, false, quot, pr, eq
Strategy
Function Precedence
p < plus = div = true = divides = zero = prime = 0 = s = times = if = false = plus# = quot = pr = eq
Argument Filtering
plus: 1 2
div: all arguments are removed from div
true: all arguments are removed from true
divides: 1 2
zero: 1
prime: all arguments are removed from prime
0: all arguments are removed from 0
s: 1
times: 1 2
if: all arguments are removed from if
p: collapses to 1
false: all arguments are removed from false
plus#: collapses to 1
quot: all arguments are removed from quot
pr: collapses to 2
eq: collapses to 2
Status
plus: lexicographic with permutation 1 → 2 2 → 1
div: multiset
true: multiset
divides: lexicographic with permutation 1 → 2 2 → 1
zero: lexicographic with permutation 1 → 1
prime: multiset
0: multiset
s: multiset
times: lexicographic with permutation 1 → 2 2 → 1
if: multiset
false: multiset
quot: multiset
Usable Rules
The dependency pairs and usable rules are stronlgy conservative!
Eliminated dependency pairs
The following dependency pairs (at least) can be eliminated according to the given precedence.
| plus#(s(x), y) → plus#(x, y) |
Problem 5: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| eq#(s(x), s(y)) | → | eq#(x, y) |
Rewrite Rules
| p(0) | → | 0 | | p(s(x)) | → | x |
| plus(x, 0) | → | x | | plus(0, y) | → | y |
| plus(s(x), y) | → | s(plus(x, y)) | | plus(s(x), y) | → | s(plus(p(s(x)), y)) |
| plus(x, s(y)) | → | s(plus(x, p(s(y)))) | | times(0, y) | → | 0 |
| times(s(0), y) | → | y | | times(s(x), y) | → | plus(y, times(x, y)) |
| div(0, y) | → | 0 | | div(x, y) | → | quot(x, y, y) |
| quot(zero(y), s(y), z) | → | 0 | | quot(s(x), s(y), z) | → | quot(x, y, z) |
| quot(x, 0, s(z)) | → | s(div(x, s(z))) | | div(div(x, y), z) | → | div(x, times(zero(y), z)) |
| eq(0, 0) | → | true | | eq(s(x), 0) | → | false |
| eq(0, s(y)) | → | false | | eq(s(x), s(y)) | → | eq(x, y) |
| divides(y, x) | → | eq(x, times(div(x, y), y)) | | prime(s(s(x))) | → | pr(s(s(x)), s(x)) |
| pr(x, s(0)) | → | true | | pr(x, s(s(y))) | → | if(divides(s(s(y)), x), x, s(y)) |
| if(true, x, y) | → | false | | if(false, x, y) | → | pr(x, y) |
| zero(div(x, x)) | → | x | | zero(divides(x, x)) | → | x |
| zero(times(x, x)) | → | x | | zero(quot(x, x, x)) | → | x |
| zero(s(x)) | → | if(eq(x, s(0)), plus(zero(0), 0), s(plus(0, zero(0)))) |
Original Signature
Termination of terms over the following signature is verified: plus, div, true, divides, zero, prime, 0, s, times, if, p, false, quot, pr, eq
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| eq#(s(x), s(y)) | → | eq#(x, y) |