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 (41ms).
| – Problem 2 was processed with processor BackwardInstantiation (2ms).
| | – Problem 5 remains open; application of the following processors failed [ForwardInstantiation (2ms), Propagation (0ms), ForwardNarrowing (1ms), BackwardInstantiation (1ms), ForwardInstantiation (1ms), Propagation (0ms)].
| – Problem 3 was processed with processor SubtermCriterion (1ms).
| – Problem 4 was processed with processor SubtermCriterion (1ms).
The following open problems remain:
Open Dependency Pair Problem 2
Dependency Pairs
| ify#(true, x, y) | → | if#(ge(x, y), x, y) | | div#(x, y) | → | ify#(ge(y, s(0)), x, y) |
| if#(true, x, y) | → | div#(minus(x, y), y) |
Rewrite Rules
| ge(x, 0) | → | true | | ge(0, s(x)) | → | false |
| ge(s(x), s(y)) | → | ge(x, y) | | minus(x, 0) | → | x |
| minus(s(x), s(y)) | → | minus(x, y) | | div(x, y) | → | ify(ge(y, s(0)), x, y) |
| ify(false, x, y) | → | divByZeroError | | ify(true, x, y) | → | if(ge(x, y), x, y) |
| if(false, x, y) | → | 0 | | if(true, x, y) | → | s(div(minus(x, y), y)) |
Original Signature
Termination of terms over the following signature is verified: minus, 0, s, if, ify, divByZeroError, div, false, true, ge
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| ify#(true, x, y) | → | if#(ge(x, y), x, y) | | ify#(true, x, y) | → | ge#(x, y) |
| div#(x, y) | → | ify#(ge(y, s(0)), x, y) | | div#(x, y) | → | ge#(y, s(0)) |
| ge#(s(x), s(y)) | → | ge#(x, y) | | minus#(s(x), s(y)) | → | minus#(x, y) |
| if#(true, x, y) | → | minus#(x, y) | | if#(true, x, y) | → | div#(minus(x, y), y) |
Rewrite Rules
| ge(x, 0) | → | true | | ge(0, s(x)) | → | false |
| ge(s(x), s(y)) | → | ge(x, y) | | minus(x, 0) | → | x |
| minus(s(x), s(y)) | → | minus(x, y) | | div(x, y) | → | ify(ge(y, s(0)), x, y) |
| ify(false, x, y) | → | divByZeroError | | ify(true, x, y) | → | if(ge(x, y), x, y) |
| if(false, x, y) | → | 0 | | if(true, x, y) | → | s(div(minus(x, y), y)) |
Original Signature
Termination of terms over the following signature is verified: 0, minus, s, ify, if, div, divByZeroError, true, false, ge
Strategy
The following SCCs where found
| ge#(s(x), s(y)) → ge#(x, y) |
| minus#(s(x), s(y)) → minus#(x, y) |
| ify#(true, x, y) → if#(ge(x, y), x, y) | div#(x, y) → ify#(ge(y, s(0)), x, y) |
| if#(true, x, y) → div#(minus(x, y), y) |
Problem 2: BackwardInstantiation
Dependency Pair Problem
Dependency Pairs
| ify#(true, x, y) | → | if#(ge(x, y), x, y) | | div#(x, y) | → | ify#(ge(y, s(0)), x, y) |
| if#(true, x, y) | → | div#(minus(x, y), y) |
Rewrite Rules
| ge(x, 0) | → | true | | ge(0, s(x)) | → | false |
| ge(s(x), s(y)) | → | ge(x, y) | | minus(x, 0) | → | x |
| minus(s(x), s(y)) | → | minus(x, y) | | div(x, y) | → | ify(ge(y, s(0)), x, y) |
| ify(false, x, y) | → | divByZeroError | | ify(true, x, y) | → | if(ge(x, y), x, y) |
| if(false, x, y) | → | 0 | | if(true, x, y) | → | s(div(minus(x, y), y)) |
Original Signature
Termination of terms over the following signature is verified: 0, minus, s, ify, if, div, divByZeroError, true, false, ge
Strategy
Instantiation
For all potential predecessors l → r of the rule div
#(
x,
y) → ify
#(ge(
y, s(0)),
x,
y) on dependency pair chains it holds that:
- div#(x, y) matches r,
- all variables of div#(x, y) are embedded in constructor contexts, i.e., each subterm of div#(x, y), containing a variable is rooted by a constructor symbol.
Thus, div
#(
x,
y) → ify
#(ge(
y, s(0)),
x,
y) is replaced by instances determined through the above matching. These instances are:
| div#(minus(_x, _y), _y) → ify#(ge(_y, s(0)), minus(_x, _y), _y) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| ge#(s(x), s(y)) | → | ge#(x, y) |
Rewrite Rules
| ge(x, 0) | → | true | | ge(0, s(x)) | → | false |
| ge(s(x), s(y)) | → | ge(x, y) | | minus(x, 0) | → | x |
| minus(s(x), s(y)) | → | minus(x, y) | | div(x, y) | → | ify(ge(y, s(0)), x, y) |
| ify(false, x, y) | → | divByZeroError | | ify(true, x, y) | → | if(ge(x, y), x, y) |
| if(false, x, y) | → | 0 | | if(true, x, y) | → | s(div(minus(x, y), y)) |
Original Signature
Termination of terms over the following signature is verified: 0, minus, s, ify, if, div, divByZeroError, true, false, ge
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| ge#(s(x), s(y)) | → | ge#(x, y) |
Problem 4: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| minus#(s(x), s(y)) | → | minus#(x, y) |
Rewrite Rules
| ge(x, 0) | → | true | | ge(0, s(x)) | → | false |
| ge(s(x), s(y)) | → | ge(x, y) | | minus(x, 0) | → | x |
| minus(s(x), s(y)) | → | minus(x, y) | | div(x, y) | → | ify(ge(y, s(0)), x, y) |
| ify(false, x, y) | → | divByZeroError | | ify(true, x, y) | → | if(ge(x, y), x, y) |
| if(false, x, y) | → | 0 | | if(true, x, y) | → | s(div(minus(x, y), y)) |
Original Signature
Termination of terms over the following signature is verified: 0, minus, s, ify, if, div, divByZeroError, true, false, ge
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| minus#(s(x), s(y)) | → | minus#(x, y) |