TIMEOUT
The TRS could not be proven terminating. The proof attempt took 60025 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (79ms).
| – 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 (1ms), DependencyGraph (8ms), PolynomialLinearRange4iUR (1224ms), DependencyGraph (7ms), PolynomialLinearRange8NegiUR (timeout), DependencyGraph (5ms), ReductionPairSAT (28646ms)].
The following open problems remain:
Open Dependency Pair Problem 5
Dependency Pairs
| if3#(true, x, y) | → | mod#(minus(x, y), s(y)) | | if2#(false, b2, x, y) | → | if3#(b2, x, y) |
| if_mod#(false, b1, b2, x, y) | → | if2#(b1, b2, x, y) | | mod#(x, y) | → | if_mod#(zero(x), zero(y), le(y, x), id(x), id(y)) |
Rewrite Rules
| le(0, y) | → | true | | le(s(x), 0) | → | false |
| le(s(x), s(y)) | → | le(x, y) | | zero(0) | → | true |
| zero(s(x)) | → | false | | id(0) | → | 0 |
| id(s(x)) | → | s(id(x)) | | minus(x, 0) | → | x |
| minus(s(x), s(y)) | → | minus(x, y) | | mod(x, y) | → | if_mod(zero(x), zero(y), le(y, x), id(x), id(y)) |
| if_mod(true, b1, b2, x, y) | → | 0 | | if_mod(false, b1, b2, x, y) | → | if2(b1, b2, x, y) |
| if2(true, b2, x, y) | → | 0 | | if2(false, b2, x, y) | → | if3(b2, x, y) |
| if3(true, x, y) | → | mod(minus(x, y), s(y)) | | if3(false, x, y) | → | x |
Original Signature
Termination of terms over the following signature is verified: id, minus, 0, s, le, mod, if3, false, true, if2, zero, if_mod
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| if3#(true, x, y) | → | mod#(minus(x, y), s(y)) | | mod#(x, y) | → | zero#(y) |
| mod#(x, y) | → | id#(x) | | id#(s(x)) | → | id#(x) |
| if3#(true, x, y) | → | minus#(x, y) | | if2#(false, b2, x, y) | → | if3#(b2, x, y) |
| le#(s(x), s(y)) | → | le#(x, y) | | mod#(x, y) | → | if_mod#(zero(x), zero(y), le(y, x), id(x), id(y)) |
| if_mod#(false, b1, b2, x, y) | → | if2#(b1, b2, x, y) | | mod#(x, y) | → | le#(y, x) |
| minus#(s(x), s(y)) | → | minus#(x, y) | | mod#(x, y) | → | zero#(x) |
| mod#(x, y) | → | id#(y) |
Rewrite Rules
| le(0, y) | → | true | | le(s(x), 0) | → | false |
| le(s(x), s(y)) | → | le(x, y) | | zero(0) | → | true |
| zero(s(x)) | → | false | | id(0) | → | 0 |
| id(s(x)) | → | s(id(x)) | | minus(x, 0) | → | x |
| minus(s(x), s(y)) | → | minus(x, y) | | mod(x, y) | → | if_mod(zero(x), zero(y), le(y, x), id(x), id(y)) |
| if_mod(true, b1, b2, x, y) | → | 0 | | if_mod(false, b1, b2, x, y) | → | if2(b1, b2, x, y) |
| if2(true, b2, x, y) | → | 0 | | if2(false, b2, x, y) | → | if3(b2, x, y) |
| if3(true, x, y) | → | mod(minus(x, y), s(y)) | | if3(false, x, y) | → | x |
Original Signature
Termination of terms over the following signature is verified: id, 0, minus, le, s, mod, if3, true, false, if2, zero, if_mod
Strategy
The following SCCs where found
| le#(s(x), s(y)) → le#(x, y) |
| minus#(s(x), s(y)) → minus#(x, y) |
| if3#(true, x, y) → mod#(minus(x, y), s(y)) | if2#(false, b2, x, y) → if3#(b2, x, y) |
| mod#(x, y) → if_mod#(zero(x), zero(y), le(y, x), id(x), id(y)) | if_mod#(false, b1, b2, x, y) → if2#(b1, b2, x, y) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| minus#(s(x), s(y)) | → | minus#(x, y) |
Rewrite Rules
| le(0, y) | → | true | | le(s(x), 0) | → | false |
| le(s(x), s(y)) | → | le(x, y) | | zero(0) | → | true |
| zero(s(x)) | → | false | | id(0) | → | 0 |
| id(s(x)) | → | s(id(x)) | | minus(x, 0) | → | x |
| minus(s(x), s(y)) | → | minus(x, y) | | mod(x, y) | → | if_mod(zero(x), zero(y), le(y, x), id(x), id(y)) |
| if_mod(true, b1, b2, x, y) | → | 0 | | if_mod(false, b1, b2, x, y) | → | if2(b1, b2, x, y) |
| if2(true, b2, x, y) | → | 0 | | if2(false, b2, x, y) | → | if3(b2, x, y) |
| if3(true, x, y) | → | mod(minus(x, y), s(y)) | | if3(false, x, y) | → | x |
Original Signature
Termination of terms over the following signature is verified: id, 0, minus, le, s, mod, if3, true, false, if2, zero, if_mod
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| minus#(s(x), s(y)) | → | minus#(x, y) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
Rewrite Rules
| le(0, y) | → | true | | le(s(x), 0) | → | false |
| le(s(x), s(y)) | → | le(x, y) | | zero(0) | → | true |
| zero(s(x)) | → | false | | id(0) | → | 0 |
| id(s(x)) | → | s(id(x)) | | minus(x, 0) | → | x |
| minus(s(x), s(y)) | → | minus(x, y) | | mod(x, y) | → | if_mod(zero(x), zero(y), le(y, x), id(x), id(y)) |
| if_mod(true, b1, b2, x, y) | → | 0 | | if_mod(false, b1, b2, x, y) | → | if2(b1, b2, x, y) |
| if2(true, b2, x, y) | → | 0 | | if2(false, b2, x, y) | → | if3(b2, x, y) |
| if3(true, x, y) | → | mod(minus(x, y), s(y)) | | if3(false, x, y) | → | x |
Original Signature
Termination of terms over the following signature is verified: id, 0, minus, le, s, mod, if3, true, false, if2, zero, if_mod
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
Problem 4: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| le#(s(x), s(y)) | → | le#(x, y) |
Rewrite Rules
| le(0, y) | → | true | | le(s(x), 0) | → | false |
| le(s(x), s(y)) | → | le(x, y) | | zero(0) | → | true |
| zero(s(x)) | → | false | | id(0) | → | 0 |
| id(s(x)) | → | s(id(x)) | | minus(x, 0) | → | x |
| minus(s(x), s(y)) | → | minus(x, y) | | mod(x, y) | → | if_mod(zero(x), zero(y), le(y, x), id(x), id(y)) |
| if_mod(true, b1, b2, x, y) | → | 0 | | if_mod(false, b1, b2, x, y) | → | if2(b1, b2, x, y) |
| if2(true, b2, x, y) | → | 0 | | if2(false, b2, x, y) | → | if3(b2, x, y) |
| if3(true, x, y) | → | mod(minus(x, y), s(y)) | | if3(false, x, y) | → | x |
Original Signature
Termination of terms over the following signature is verified: id, 0, minus, le, s, mod, if3, true, false, if2, zero, if_mod
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| le#(s(x), s(y)) | → | le#(x, y) |