TIMEOUT
The TRS could not be proven terminating. The proof attempt took 60001 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (67ms).
| – Problem 2 was processed with processor SubtermCriterion (1ms).
| – Problem 3 was processed with processor SubtermCriterion (0ms).
| – Problem 4 was processed with processor BackwardInstantiation (2ms).
| | – Problem 6 remains open; application of the following processors failed [ForwardInstantiation (1ms), Propagation (1ms), ForwardNarrowing (1ms), BackwardInstantiation (1ms), ForwardInstantiation (1ms), Propagation (3ms)].
| – Problem 5 was processed with processor SubtermCriterion (0ms).
The following open problems remain:
Open Dependency Pair Problem 4
Dependency Pairs
| if#(true, x, s(y), c) | → | help#(x, s(y), plus(c, s(y))) | | help#(x, s(y), c) | → | if#(le(c, x), x, s(y), c) |
Rewrite Rules
| le(0, y) | → | true | | le(s(x), 0) | → | false |
| le(s(x), s(y)) | → | le(x, y) | | minus(x, 0) | → | x |
| minus(0, s(y)) | → | 0 | | minus(s(x), s(y)) | → | minus(x, y) |
| plus(x, 0) | → | x | | plus(x, s(y)) | → | s(plus(x, y)) |
| mod(s(x), 0) | → | 0 | | mod(x, s(y)) | → | help(x, s(y), 0) |
| help(x, s(y), c) | → | if(le(c, x), x, s(y), c) | | if(true, x, s(y), c) | → | help(x, s(y), plus(c, s(y))) |
| if(false, x, s(y), c) | → | minus(x, minus(c, s(y))) |
Original Signature
Termination of terms over the following signature is verified: plus, help, minus, 0, s, le, if, mod, false, true
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| le#(s(x), s(y)) | → | le#(x, y) | | plus#(x, s(y)) | → | plus#(x, y) |
| mod#(x, s(y)) | → | help#(x, s(y), 0) | | minus#(s(x), s(y)) | → | minus#(x, y) |
| if#(true, x, s(y), c) | → | help#(x, s(y), plus(c, s(y))) | | if#(false, x, s(y), c) | → | minus#(x, minus(c, s(y))) |
| help#(x, s(y), c) | → | if#(le(c, x), x, s(y), c) | | if#(false, x, s(y), c) | → | minus#(c, s(y)) |
| help#(x, s(y), c) | → | le#(c, x) | | if#(true, x, s(y), c) | → | plus#(c, s(y)) |
Rewrite Rules
| le(0, y) | → | true | | le(s(x), 0) | → | false |
| le(s(x), s(y)) | → | le(x, y) | | minus(x, 0) | → | x |
| minus(0, s(y)) | → | 0 | | minus(s(x), s(y)) | → | minus(x, y) |
| plus(x, 0) | → | x | | plus(x, s(y)) | → | s(plus(x, y)) |
| mod(s(x), 0) | → | 0 | | mod(x, s(y)) | → | help(x, s(y), 0) |
| help(x, s(y), c) | → | if(le(c, x), x, s(y), c) | | if(true, x, s(y), c) | → | help(x, s(y), plus(c, s(y))) |
| if(false, x, s(y), c) | → | minus(x, minus(c, s(y))) |
Original Signature
Termination of terms over the following signature is verified: plus, help, 0, minus, le, s, if, mod, true, false
Strategy
The following SCCs where found
| le#(s(x), s(y)) → le#(x, y) |
| if#(true, x, s(y), c) → help#(x, s(y), plus(c, s(y))) | help#(x, s(y), c) → if#(le(c, x), x, s(y), c) |
| plus#(x, s(y)) → plus#(x, y) |
| minus#(s(x), s(y)) → minus#(x, y) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| plus#(x, s(y)) | → | plus#(x, y) |
Rewrite Rules
| le(0, y) | → | true | | le(s(x), 0) | → | false |
| le(s(x), s(y)) | → | le(x, y) | | minus(x, 0) | → | x |
| minus(0, s(y)) | → | 0 | | minus(s(x), s(y)) | → | minus(x, y) |
| plus(x, 0) | → | x | | plus(x, s(y)) | → | s(plus(x, y)) |
| mod(s(x), 0) | → | 0 | | mod(x, s(y)) | → | help(x, s(y), 0) |
| help(x, s(y), c) | → | if(le(c, x), x, s(y), c) | | if(true, x, s(y), c) | → | help(x, s(y), plus(c, s(y))) |
| if(false, x, s(y), c) | → | minus(x, minus(c, s(y))) |
Original Signature
Termination of terms over the following signature is verified: plus, help, 0, minus, le, s, if, mod, true, false
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| plus#(x, s(y)) | → | plus#(x, y) |
Problem 3: 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) | | minus(x, 0) | → | x |
| minus(0, s(y)) | → | 0 | | minus(s(x), s(y)) | → | minus(x, y) |
| plus(x, 0) | → | x | | plus(x, s(y)) | → | s(plus(x, y)) |
| mod(s(x), 0) | → | 0 | | mod(x, s(y)) | → | help(x, s(y), 0) |
| help(x, s(y), c) | → | if(le(c, x), x, s(y), c) | | if(true, x, s(y), c) | → | help(x, s(y), plus(c, s(y))) |
| if(false, x, s(y), c) | → | minus(x, minus(c, s(y))) |
Original Signature
Termination of terms over the following signature is verified: plus, help, 0, minus, le, s, if, mod, true, false
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| le#(s(x), s(y)) | → | le#(x, y) |
Problem 4: BackwardInstantiation
Dependency Pair Problem
Dependency Pairs
| if#(true, x, s(y), c) | → | help#(x, s(y), plus(c, s(y))) | | help#(x, s(y), c) | → | if#(le(c, x), x, s(y), c) |
Rewrite Rules
| le(0, y) | → | true | | le(s(x), 0) | → | false |
| le(s(x), s(y)) | → | le(x, y) | | minus(x, 0) | → | x |
| minus(0, s(y)) | → | 0 | | minus(s(x), s(y)) | → | minus(x, y) |
| plus(x, 0) | → | x | | plus(x, s(y)) | → | s(plus(x, y)) |
| mod(s(x), 0) | → | 0 | | mod(x, s(y)) | → | help(x, s(y), 0) |
| help(x, s(y), c) | → | if(le(c, x), x, s(y), c) | | if(true, x, s(y), c) | → | help(x, s(y), plus(c, s(y))) |
| if(false, x, s(y), c) | → | minus(x, minus(c, s(y))) |
Original Signature
Termination of terms over the following signature is verified: plus, help, 0, minus, le, s, if, mod, true, false
Strategy
Instantiation
For all potential predecessors l → r of the rule help
#(
x, s(
y),
c) → if
#(le(
c,
x),
x, s(
y),
c) on dependency pair chains it holds that:
- help#(x, s(y), c) matches r,
- all variables of help#(x, s(y), c) are embedded in constructor contexts, i.e., each subterm of help#(x, s(y), c), containing a variable is rooted by a constructor symbol.
Thus, help
#(
x, s(
y),
c) → if
#(le(
c,
x),
x, s(
y),
c) is replaced by instances determined through the above matching. These instances are:
| help#(_x, s(_y), plus(_c, s(_y))) → if#(le(plus(_c, s(_y)), _x), _x, s(_y), plus(_c, s(_y))) |
Problem 5: 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) | | minus(x, 0) | → | x |
| minus(0, s(y)) | → | 0 | | minus(s(x), s(y)) | → | minus(x, y) |
| plus(x, 0) | → | x | | plus(x, s(y)) | → | s(plus(x, y)) |
| mod(s(x), 0) | → | 0 | | mod(x, s(y)) | → | help(x, s(y), 0) |
| help(x, s(y), c) | → | if(le(c, x), x, s(y), c) | | if(true, x, s(y), c) | → | help(x, s(y), plus(c, s(y))) |
| if(false, x, s(y), c) | → | minus(x, minus(c, s(y))) |
Original Signature
Termination of terms over the following signature is verified: plus, help, 0, minus, le, s, if, mod, true, false
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| minus#(s(x), s(y)) | → | minus#(x, y) |