TIMEOUT
The TRS could not be proven terminating. The proof attempt took 60024 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (37ms).
| – Problem 2 was processed with processor BackwardInstantiation (3ms).
| | – Problem 5 remains open; application of the following processors failed [ForwardInstantiation (2ms), Propagation (1ms), ForwardNarrowing (0ms), BackwardInstantiation (1ms), ForwardInstantiation (2ms), Propagation (1ms)].
| – Problem 3 was processed with processor SubtermCriterion (1ms).
| – Problem 4 was processed with processor SubtermCriterion (0ms).
The following open problems remain:
Open Dependency Pair Problem 2
Dependency Pairs
| if#(true, x, s(y), c) | → | help#(x, s(y), plus(c, s(y))) | | help#(x, s(y), c) | → | if#(lt(c, x), x, s(y), c) |
Rewrite Rules
| lt(x, 0) | → | false | | lt(0, s(y)) | → | true |
| lt(s(x), s(y)) | → | lt(x, y) | | plus(x, 0) | → | x |
| plus(x, s(y)) | → | s(plus(x, y)) | | quot(x, s(y)) | → | help(x, s(y), 0) |
| help(x, s(y), c) | → | if(lt(c, x), x, s(y), c) | | if(true, x, s(y), c) | → | s(help(x, s(y), plus(c, s(y)))) |
| if(false, x, s(y), c) | → | 0 |
Original Signature
Termination of terms over the following signature is verified: plus, help, 0, s, if, true, false, lt, quot
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| quot#(x, s(y)) | → | help#(x, s(y), 0) | | plus#(x, s(y)) | → | plus#(x, y) |
| help#(x, s(y), c) | → | lt#(c, x) | | if#(true, x, s(y), c) | → | help#(x, s(y), plus(c, s(y))) |
| lt#(s(x), s(y)) | → | lt#(x, y) | | help#(x, s(y), c) | → | if#(lt(c, x), x, s(y), c) |
| if#(true, x, s(y), c) | → | plus#(c, s(y)) |
Rewrite Rules
| lt(x, 0) | → | false | | lt(0, s(y)) | → | true |
| lt(s(x), s(y)) | → | lt(x, y) | | plus(x, 0) | → | x |
| plus(x, s(y)) | → | s(plus(x, y)) | | quot(x, s(y)) | → | help(x, s(y), 0) |
| help(x, s(y), c) | → | if(lt(c, x), x, s(y), c) | | if(true, x, s(y), c) | → | s(help(x, s(y), plus(c, s(y)))) |
| if(false, x, s(y), c) | → | 0 |
Original Signature
Termination of terms over the following signature is verified: plus, help, 0, s, if, false, true, lt, quot
Strategy
The following SCCs where found
| plus#(x, s(y)) → plus#(x, y) |
| if#(true, x, s(y), c) → help#(x, s(y), plus(c, s(y))) | help#(x, s(y), c) → if#(lt(c, x), x, s(y), c) |
| lt#(s(x), s(y)) → lt#(x, y) |
Problem 2: 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#(lt(c, x), x, s(y), c) |
Rewrite Rules
| lt(x, 0) | → | false | | lt(0, s(y)) | → | true |
| lt(s(x), s(y)) | → | lt(x, y) | | plus(x, 0) | → | x |
| plus(x, s(y)) | → | s(plus(x, y)) | | quot(x, s(y)) | → | help(x, s(y), 0) |
| help(x, s(y), c) | → | if(lt(c, x), x, s(y), c) | | if(true, x, s(y), c) | → | s(help(x, s(y), plus(c, s(y)))) |
| if(false, x, s(y), c) | → | 0 |
Original Signature
Termination of terms over the following signature is verified: plus, help, 0, s, if, false, true, lt, quot
Strategy
Instantiation
For all potential predecessors l → r of the rule help
#(
x, s(
y),
c) → if
#(lt(
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
#(lt(
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#(lt(plus(_c, s(_y)), _x), _x, s(_y), plus(_c, s(_y))) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| lt#(s(x), s(y)) | → | lt#(x, y) |
Rewrite Rules
| lt(x, 0) | → | false | | lt(0, s(y)) | → | true |
| lt(s(x), s(y)) | → | lt(x, y) | | plus(x, 0) | → | x |
| plus(x, s(y)) | → | s(plus(x, y)) | | quot(x, s(y)) | → | help(x, s(y), 0) |
| help(x, s(y), c) | → | if(lt(c, x), x, s(y), c) | | if(true, x, s(y), c) | → | s(help(x, s(y), plus(c, s(y)))) |
| if(false, x, s(y), c) | → | 0 |
Original Signature
Termination of terms over the following signature is verified: plus, help, 0, s, if, false, true, lt, quot
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| lt#(s(x), s(y)) | → | lt#(x, y) |
Problem 4: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| plus#(x, s(y)) | → | plus#(x, y) |
Rewrite Rules
| lt(x, 0) | → | false | | lt(0, s(y)) | → | true |
| lt(s(x), s(y)) | → | lt(x, y) | | plus(x, 0) | → | x |
| plus(x, s(y)) | → | s(plus(x, y)) | | quot(x, s(y)) | → | help(x, s(y), 0) |
| help(x, s(y), c) | → | if(lt(c, x), x, s(y), c) | | if(true, x, s(y), c) | → | s(help(x, s(y), plus(c, s(y)))) |
| if(false, x, s(y), c) | → | 0 |
Original Signature
Termination of terms over the following signature is verified: plus, help, 0, s, if, false, true, lt, quot
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| plus#(x, s(y)) | → | plus#(x, y) |