TIMEOUT
The TRS could not be proven terminating. The proof attempt took 60079 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (48ms).
| – Problem 2 was processed with processor SubtermCriterion (2ms).
| – Problem 3 was processed with processor SubtermCriterion (0ms).
| – Problem 4 was processed with processor BackwardInstantiation (4ms).
| | – Problem 6 remains open; application of the following processors failed [ForwardInstantiation (2ms), Propagation (0ms), ForwardNarrowing (3ms), BackwardInstantiation (1ms), ForwardInstantiation (1ms), Propagation (1ms)].
| – Problem 5 was processed with processor SubtermCriterion (1ms).
The following open problems remain:
Open Dependency Pair Problem 4
Dependency Pairs
| if#(false, x, s(y), z) | → | div#(minus(x, s(y)), s(y), z) | | div#(x, y, z) | → | if#(lt(x, y), x, y, inc(z)) |
Rewrite Rules
| division(x, y) | → | div(x, y, 0) | | div(x, y, z) | → | if(lt(x, y), x, y, inc(z)) |
| if(true, x, y, z) | → | z | | if(false, x, s(y), z) | → | div(minus(x, s(y)), s(y), z) |
| minus(x, 0) | → | x | | minus(s(x), s(y)) | → | minus(x, y) |
| lt(x, 0) | → | false | | lt(0, s(y)) | → | true |
| lt(s(x), s(y)) | → | lt(x, y) | | inc(0) | → | s(0) |
| inc(s(x)) | → | s(inc(x)) |
Original Signature
Termination of terms over the following signature is verified: minus, division, 0, s, inc, if, div, false, true, lt
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| div#(x, y, z) | → | lt#(x, y) | | division#(x, y) | → | div#(x, y, 0) |
| if#(false, x, s(y), z) | → | div#(minus(x, s(y)), s(y), z) | | if#(false, x, s(y), z) | → | minus#(x, s(y)) |
| minus#(s(x), s(y)) | → | minus#(x, y) | | div#(x, y, z) | → | inc#(z) |
| lt#(s(x), s(y)) | → | lt#(x, y) | | inc#(s(x)) | → | inc#(x) |
| div#(x, y, z) | → | if#(lt(x, y), x, y, inc(z)) |
Rewrite Rules
| division(x, y) | → | div(x, y, 0) | | div(x, y, z) | → | if(lt(x, y), x, y, inc(z)) |
| if(true, x, y, z) | → | z | | if(false, x, s(y), z) | → | div(minus(x, s(y)), s(y), z) |
| minus(x, 0) | → | x | | minus(s(x), s(y)) | → | minus(x, y) |
| lt(x, 0) | → | false | | lt(0, s(y)) | → | true |
| lt(s(x), s(y)) | → | lt(x, y) | | inc(0) | → | s(0) |
| inc(s(x)) | → | s(inc(x)) |
Original Signature
Termination of terms over the following signature is verified: 0, division, minus, inc, s, if, div, true, false, lt
Strategy
The following SCCs where found
| if#(false, x, s(y), z) → div#(minus(x, s(y)), s(y), z) | div#(x, y, z) → if#(lt(x, y), x, y, inc(z)) |
| minus#(s(x), s(y)) → minus#(x, y) |
| lt#(s(x), s(y)) → lt#(x, y) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
Rewrite Rules
| division(x, y) | → | div(x, y, 0) | | div(x, y, z) | → | if(lt(x, y), x, y, inc(z)) |
| if(true, x, y, z) | → | z | | if(false, x, s(y), z) | → | div(minus(x, s(y)), s(y), z) |
| minus(x, 0) | → | x | | minus(s(x), s(y)) | → | minus(x, y) |
| lt(x, 0) | → | false | | lt(0, s(y)) | → | true |
| lt(s(x), s(y)) | → | lt(x, y) | | inc(0) | → | s(0) |
| inc(s(x)) | → | s(inc(x)) |
Original Signature
Termination of terms over the following signature is verified: 0, division, minus, inc, s, if, div, true, false, lt
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| lt#(s(x), s(y)) | → | lt#(x, y) |
Rewrite Rules
| division(x, y) | → | div(x, y, 0) | | div(x, y, z) | → | if(lt(x, y), x, y, inc(z)) |
| if(true, x, y, z) | → | z | | if(false, x, s(y), z) | → | div(minus(x, s(y)), s(y), z) |
| minus(x, 0) | → | x | | minus(s(x), s(y)) | → | minus(x, y) |
| lt(x, 0) | → | false | | lt(0, s(y)) | → | true |
| lt(s(x), s(y)) | → | lt(x, y) | | inc(0) | → | s(0) |
| inc(s(x)) | → | s(inc(x)) |
Original Signature
Termination of terms over the following signature is verified: 0, division, minus, inc, s, if, div, true, false, lt
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| lt#(s(x), s(y)) | → | lt#(x, y) |
Problem 4: BackwardInstantiation
Dependency Pair Problem
Dependency Pairs
| if#(false, x, s(y), z) | → | div#(minus(x, s(y)), s(y), z) | | div#(x, y, z) | → | if#(lt(x, y), x, y, inc(z)) |
Rewrite Rules
| division(x, y) | → | div(x, y, 0) | | div(x, y, z) | → | if(lt(x, y), x, y, inc(z)) |
| if(true, x, y, z) | → | z | | if(false, x, s(y), z) | → | div(minus(x, s(y)), s(y), z) |
| minus(x, 0) | → | x | | minus(s(x), s(y)) | → | minus(x, y) |
| lt(x, 0) | → | false | | lt(0, s(y)) | → | true |
| lt(s(x), s(y)) | → | lt(x, y) | | inc(0) | → | s(0) |
| inc(s(x)) | → | s(inc(x)) |
Original Signature
Termination of terms over the following signature is verified: 0, division, minus, inc, s, if, div, true, false, lt
Strategy
Instantiation
For all potential predecessors l → r of the rule div
#(
x,
y,
z) → if
#(lt(
x,
y),
x,
y, inc(
z)) on dependency pair chains it holds that:
- div#(x, y, z) matches r,
- all variables of div#(x, y, z) are embedded in constructor contexts, i.e., each subterm of div#(x, y, z), containing a variable is rooted by a constructor symbol.
Thus, div
#(
x,
y,
z) → if
#(lt(
x,
y),
x,
y, inc(
z)) is replaced by instances determined through the above matching. These instances are:
| div#(minus(_x, s(_y)), s(_y), _z) → if#(lt(minus(_x, s(_y)), s(_y)), minus(_x, s(_y)), s(_y), inc(_z)) |
Problem 5: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| minus#(s(x), s(y)) | → | minus#(x, y) |
Rewrite Rules
| division(x, y) | → | div(x, y, 0) | | div(x, y, z) | → | if(lt(x, y), x, y, inc(z)) |
| if(true, x, y, z) | → | z | | if(false, x, s(y), z) | → | div(minus(x, s(y)), s(y), z) |
| minus(x, 0) | → | x | | minus(s(x), s(y)) | → | minus(x, y) |
| lt(x, 0) | → | false | | lt(0, s(y)) | → | true |
| lt(s(x), s(y)) | → | lt(x, y) | | inc(0) | → | s(0) |
| inc(s(x)) | → | s(inc(x)) |
Original Signature
Termination of terms over the following signature is verified: 0, division, minus, inc, s, if, div, true, false, lt
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| minus#(s(x), s(y)) | → | minus#(x, y) |