TIMEOUT
The TRS could not be proven terminating. The proof attempt took 60045 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (206ms).
| – Problem 2 was processed with processor SubtermCriterion (1ms).
| – Problem 3 remains open; application of the following processors failed [SubtermCriterion (1ms), DependencyGraph (5ms), PolynomialLinearRange4iUR (1287ms), DependencyGraph (4ms), PolynomialLinearRange8NegiUR (5376ms), DependencyGraph (4ms), ReductionPairSAT (1444ms)].
| – Problem 4 was processed with processor SubtermCriterion (0ms).
| – Problem 5 remains open; application of the following processors failed [SubtermCriterion (0ms), DependencyGraph (4ms), PolynomialLinearRange4iUR (5001ms), DependencyGraph (4ms), PolynomialLinearRange8NegiUR (15001ms), DependencyGraph (3ms), ReductionPairSAT (timeout)].
The following open problems remain:
Open Dependency Pair Problem 3
Dependency Pairs
| plus#(id(x), s(y)) | → | plus#(x, if(gt(s(y), y), y, s(y))) | | plus#(s(x), s(y)) | → | plus#(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y))) |
| plus#(s(x), x) | → | plus#(if(gt(x, x), id(x), id(x)), s(x)) |
Rewrite Rules
| app(nil, k) | → | k | | app(l, nil) | → | l |
| app(cons(x, l), k) | → | cons(x, app(l, k)) | | sum(cons(x, nil)) | → | cons(x, nil) |
| sum(cons(x, cons(y, l))) | → | sum(cons(plus(x, y), l)) | | sum(app(l, cons(x, cons(y, k)))) | → | sum(app(l, sum(cons(x, cons(y, k))))) |
| plus(s(x), s(y)) | → | s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y))))) | | plus(s(x), x) | → | plus(if(gt(x, x), id(x), id(x)), s(x)) |
| plus(zero, y) | → | y | | plus(id(x), s(y)) | → | s(plus(x, if(gt(s(y), y), y, s(y)))) |
| id(x) | → | x | | if(true, x, y) | → | x |
| if(false, x, y) | → | y | | not(x) | → | if(x, false, true) |
| gt(s(x), zero) | → | true | | gt(zero, y) | → | false |
| gt(s(x), s(y)) | → | gt(x, y) |
Original Signature
Termination of terms over the following signature is verified: app, plus, sum, true, zero, id, not, s, if, false, gt, nil, cons
Open Dependency Pair Problem 5
Dependency Pairs
| sum#(cons(x, cons(y, l))) | → | sum#(cons(plus(x, y), l)) | | sum#(app(l, cons(x, cons(y, k)))) | → | sum#(app(l, sum(cons(x, cons(y, k))))) |
| sum#(app(l, cons(x, cons(y, k)))) | → | sum#(cons(x, cons(y, k))) |
Rewrite Rules
| app(nil, k) | → | k | | app(l, nil) | → | l |
| app(cons(x, l), k) | → | cons(x, app(l, k)) | | sum(cons(x, nil)) | → | cons(x, nil) |
| sum(cons(x, cons(y, l))) | → | sum(cons(plus(x, y), l)) | | sum(app(l, cons(x, cons(y, k)))) | → | sum(app(l, sum(cons(x, cons(y, k))))) |
| plus(s(x), s(y)) | → | s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y))))) | | plus(s(x), x) | → | plus(if(gt(x, x), id(x), id(x)), s(x)) |
| plus(zero, y) | → | y | | plus(id(x), s(y)) | → | s(plus(x, if(gt(s(y), y), y, s(y)))) |
| id(x) | → | x | | if(true, x, y) | → | x |
| if(false, x, y) | → | y | | not(x) | → | if(x, false, true) |
| gt(s(x), zero) | → | true | | gt(zero, y) | → | false |
| gt(s(x), s(y)) | → | gt(x, y) |
Original Signature
Termination of terms over the following signature is verified: app, plus, sum, true, zero, id, not, s, if, false, gt, nil, cons
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| plus#(s(x), x) | → | id#(x) | | plus#(id(x), s(y)) | → | plus#(x, if(gt(s(y), y), y, s(y))) |
| sum#(cons(x, cons(y, l))) | → | sum#(cons(plus(x, y), l)) | | plus#(id(x), s(y)) | → | gt#(s(y), y) |
| plus#(s(x), s(y)) | → | gt#(x, y) | | plus#(s(x), x) | → | plus#(if(gt(x, x), id(x), id(x)), s(x)) |
| plus#(s(x), s(y)) | → | if#(gt(x, y), x, y) | | plus#(s(x), x) | → | gt#(x, x) |
| plus#(id(x), s(y)) | → | if#(gt(s(y), y), y, s(y)) | | plus#(s(x), x) | → | if#(gt(x, x), id(x), id(x)) |
| plus#(s(x), s(y)) | → | id#(x) | | sum#(app(l, cons(x, cons(y, k)))) | → | app#(l, sum(cons(x, cons(y, k)))) |
| plus#(s(x), s(y)) | → | plus#(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y))) | | sum#(cons(x, cons(y, l))) | → | plus#(x, y) |
| plus#(s(x), s(y)) | → | if#(not(gt(x, y)), id(x), id(y)) | | sum#(app(l, cons(x, cons(y, k)))) | → | sum#(cons(x, cons(y, k))) |
| sum#(app(l, cons(x, cons(y, k)))) | → | sum#(app(l, sum(cons(x, cons(y, k))))) | | gt#(s(x), s(y)) | → | gt#(x, y) |
| not#(x) | → | if#(x, false, true) | | plus#(s(x), s(y)) | → | id#(y) |
| plus#(s(x), s(y)) | → | not#(gt(x, y)) | | app#(cons(x, l), k) | → | app#(l, k) |
Rewrite Rules
| app(nil, k) | → | k | | app(l, nil) | → | l |
| app(cons(x, l), k) | → | cons(x, app(l, k)) | | sum(cons(x, nil)) | → | cons(x, nil) |
| sum(cons(x, cons(y, l))) | → | sum(cons(plus(x, y), l)) | | sum(app(l, cons(x, cons(y, k)))) | → | sum(app(l, sum(cons(x, cons(y, k))))) |
| plus(s(x), s(y)) | → | s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y))))) | | plus(s(x), x) | → | plus(if(gt(x, x), id(x), id(x)), s(x)) |
| plus(zero, y) | → | y | | plus(id(x), s(y)) | → | s(plus(x, if(gt(s(y), y), y, s(y)))) |
| id(x) | → | x | | if(true, x, y) | → | x |
| if(false, x, y) | → | y | | not(x) | → | if(x, false, true) |
| gt(s(x), zero) | → | true | | gt(zero, y) | → | false |
| gt(s(x), s(y)) | → | gt(x, y) |
Original Signature
Termination of terms over the following signature is verified: app, plus, sum, true, zero, id, not, s, if, false, gt, nil, cons
Strategy
The following SCCs where found
| sum#(cons(x, cons(y, l))) → sum#(cons(plus(x, y), l)) | sum#(app(l, cons(x, cons(y, k)))) → sum#(cons(x, cons(y, k))) |
| sum#(app(l, cons(x, cons(y, k)))) → sum#(app(l, sum(cons(x, cons(y, k))))) |
| plus#(id(x), s(y)) → plus#(x, if(gt(s(y), y), y, s(y))) | plus#(s(x), s(y)) → plus#(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y))) |
| plus#(s(x), x) → plus#(if(gt(x, x), id(x), id(x)), s(x)) |
| gt#(s(x), s(y)) → gt#(x, y) |
| app#(cons(x, l), k) → app#(l, k) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| app#(cons(x, l), k) | → | app#(l, k) |
Rewrite Rules
| app(nil, k) | → | k | | app(l, nil) | → | l |
| app(cons(x, l), k) | → | cons(x, app(l, k)) | | sum(cons(x, nil)) | → | cons(x, nil) |
| sum(cons(x, cons(y, l))) | → | sum(cons(plus(x, y), l)) | | sum(app(l, cons(x, cons(y, k)))) | → | sum(app(l, sum(cons(x, cons(y, k))))) |
| plus(s(x), s(y)) | → | s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y))))) | | plus(s(x), x) | → | plus(if(gt(x, x), id(x), id(x)), s(x)) |
| plus(zero, y) | → | y | | plus(id(x), s(y)) | → | s(plus(x, if(gt(s(y), y), y, s(y)))) |
| id(x) | → | x | | if(true, x, y) | → | x |
| if(false, x, y) | → | y | | not(x) | → | if(x, false, true) |
| gt(s(x), zero) | → | true | | gt(zero, y) | → | false |
| gt(s(x), s(y)) | → | gt(x, y) |
Original Signature
Termination of terms over the following signature is verified: app, plus, sum, true, zero, id, not, s, if, false, gt, nil, cons
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| app#(cons(x, l), k) | → | app#(l, k) |
Problem 4: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| gt#(s(x), s(y)) | → | gt#(x, y) |
Rewrite Rules
| app(nil, k) | → | k | | app(l, nil) | → | l |
| app(cons(x, l), k) | → | cons(x, app(l, k)) | | sum(cons(x, nil)) | → | cons(x, nil) |
| sum(cons(x, cons(y, l))) | → | sum(cons(plus(x, y), l)) | | sum(app(l, cons(x, cons(y, k)))) | → | sum(app(l, sum(cons(x, cons(y, k))))) |
| plus(s(x), s(y)) | → | s(s(plus(if(gt(x, y), x, y), if(not(gt(x, y)), id(x), id(y))))) | | plus(s(x), x) | → | plus(if(gt(x, x), id(x), id(x)), s(x)) |
| plus(zero, y) | → | y | | plus(id(x), s(y)) | → | s(plus(x, if(gt(s(y), y), y, s(y)))) |
| id(x) | → | x | | if(true, x, y) | → | x |
| if(false, x, y) | → | y | | not(x) | → | if(x, false, true) |
| gt(s(x), zero) | → | true | | gt(zero, y) | → | false |
| gt(s(x), s(y)) | → | gt(x, y) |
Original Signature
Termination of terms over the following signature is verified: app, plus, sum, true, zero, id, not, s, if, false, gt, nil, cons
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| gt#(s(x), s(y)) | → | gt#(x, y) |