TIMEOUT
The TRS could not be proven terminating. The proof attempt took 60017 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (29ms).
| – Problem 2 was processed with processor BackwardInstantiation (4ms).
| | – Problem 4 was processed with processor BackwardInstantiation (1ms).
| | | – Problem 5 was processed with processor Propagation (2ms).
| | | | – Problem 6 remains open; application of the following processors failed [ForwardNarrowing (0ms), BackwardInstantiation (1ms), ForwardInstantiation (1ms), Propagation (1ms)].
| – Problem 3 was processed with processor SubtermCriterion (1ms).
The following open problems remain:
Open Dependency Pair Problem 2
Dependency Pairs
| help#(true, x, y) | → | minus#(x, s(y)) | | minus#(x, y) | → | help#(lt(y, x), x, y) |
Rewrite Rules
| lt(0, s(x)) | → | true | | lt(x, 0) | → | false |
| lt(s(x), s(y)) | → | lt(x, y) | | minus(x, y) | → | help(lt(y, x), x, y) |
| help(true, x, y) | → | s(minus(x, s(y))) | | help(false, x, y) | → | 0 |
Original Signature
Termination of terms over the following signature is verified: help, minus, 0, s, false, true, lt
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| help#(true, x, y) | → | minus#(x, s(y)) | | minus#(x, y) | → | help#(lt(y, x), x, y) |
| minus#(x, y) | → | lt#(y, x) | | lt#(s(x), s(y)) | → | lt#(x, y) |
Rewrite Rules
| lt(0, s(x)) | → | true | | lt(x, 0) | → | false |
| lt(s(x), s(y)) | → | lt(x, y) | | minus(x, y) | → | help(lt(y, x), x, y) |
| help(true, x, y) | → | s(minus(x, s(y))) | | help(false, x, y) | → | 0 |
Original Signature
Termination of terms over the following signature is verified: help, 0, minus, s, true, false, lt
Strategy
The following SCCs where found
| help#(true, x, y) → minus#(x, s(y)) | minus#(x, y) → help#(lt(y, x), x, y) |
| lt#(s(x), s(y)) → lt#(x, y) |
Problem 2: BackwardInstantiation
Dependency Pair Problem
Dependency Pairs
| help#(true, x, y) | → | minus#(x, s(y)) | | minus#(x, y) | → | help#(lt(y, x), x, y) |
Rewrite Rules
| lt(0, s(x)) | → | true | | lt(x, 0) | → | false |
| lt(s(x), s(y)) | → | lt(x, y) | | minus(x, y) | → | help(lt(y, x), x, y) |
| help(true, x, y) | → | s(minus(x, s(y))) | | help(false, x, y) | → | 0 |
Original Signature
Termination of terms over the following signature is verified: help, 0, minus, s, true, false, lt
Strategy
Instantiation
For all potential predecessors l → r of the rule minus
#(
x,
y) → help
#(lt(
y,
x),
x,
y) on dependency pair chains it holds that:
- minus#(x, y) matches r,
- all variables of minus#(x, y) are embedded in constructor contexts, i.e., each subterm of minus#(x, y), containing a variable is rooted by a constructor symbol.
Thus, minus
#(
x,
y) → help
#(lt(
y,
x),
x,
y) is replaced by instances determined through the above matching. These instances are:
| minus#(_x, s(_y)) → help#(lt(s(_y), _x), _x, s(_y)) |
Problem 4: BackwardInstantiation
Dependency Pair Problem
Dependency Pairs
| help#(true, x, y) | → | minus#(x, s(y)) | | minus#(_x, s(_y)) | → | help#(lt(s(_y), _x), _x, s(_y)) |
Rewrite Rules
| lt(0, s(x)) | → | true | | lt(x, 0) | → | false |
| lt(s(x), s(y)) | → | lt(x, y) | | minus(x, y) | → | help(lt(y, x), x, y) |
| help(true, x, y) | → | s(minus(x, s(y))) | | help(false, x, y) | → | 0 |
Original Signature
Termination of terms over the following signature is verified: help, minus, 0, s, false, true, lt
Strategy
Instantiation
For all potential predecessors l → r of the rule minus
#(
_x, s(
_y)) → help
#(lt(s(
_y),
_x),
_x, s(
_y)) on dependency pair chains it holds that:
- minus#(_x, s(_y)) matches r,
- all variables of minus#(_x, s(_y)) are embedded in constructor contexts, i.e., each subterm of minus#(_x, s(_y)), containing a variable is rooted by a constructor symbol.
Thus, minus
#(
_x, s(
_y)) → help
#(lt(s(
_y),
_x),
_x, s(
_y)) is replaced by instances determined through the above matching. These instances are:
| minus#(x, s(y)) → help#(lt(s(y), x), x, s(y)) |
Problem 5: Propagation
Dependency Pair Problem
Dependency Pairs
| help#(true, x, y) | → | minus#(x, s(y)) | | minus#(x, s(y)) | → | help#(lt(s(y), x), x, s(y)) |
Rewrite Rules
| lt(0, s(x)) | → | true | | lt(x, 0) | → | false |
| lt(s(x), s(y)) | → | lt(x, y) | | minus(x, y) | → | help(lt(y, x), x, y) |
| help(true, x, y) | → | s(minus(x, s(y))) | | help(false, x, y) | → | 0 |
Original Signature
Termination of terms over the following signature is verified: help, 0, minus, s, true, false, lt
Strategy
The dependency pairs help
#(true,
x,
y) → minus
#(
x, s(
y)) and minus
#(
x, s(
y)) → help
#(lt(s(
y),
x),
x, s(
y)) are consolidated into the rule help
#(true,
x,
y) → help
#(lt(s(
y),
x),
x, s(
y)) .
This is possible as
- all subterms of minus#(x, s(y)) containing variables are rooted by a constructor symbol,
- there is no variable that is replacing in minus#(x, s(y)), but non-replacing in both help#(true, x, y) and help#(lt(s(y), x), x, s(y))
The dependency pairs help
#(true,
x,
y) → minus
#(
x, s(
y)) and minus
#(
x, s(
y)) → help
#(lt(s(
y),
x),
x, s(
y)) are consolidated into the rule help
#(true,
x,
y) → help
#(lt(s(
y),
x),
x, s(
y)) .
This is possible as
- all subterms of minus#(x, s(y)) containing variables are rooted by a constructor symbol,
- there is no variable that is replacing in minus#(x, s(y)), but non-replacing in both help#(true, x, y) and help#(lt(s(y), x), x, s(y))
Summary
| Removed Dependency Pairs | Added Dependency Pairs |
|---|
| help#(true, x, y) → minus#(x, s(y)) | help#(true, x, y) → help#(lt(s(y), x), x, s(y)) |
| minus#(x, s(y)) → help#(lt(s(y), x), x, s(y)) | |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| lt#(s(x), s(y)) | → | lt#(x, y) |
Rewrite Rules
| lt(0, s(x)) | → | true | | lt(x, 0) | → | false |
| lt(s(x), s(y)) | → | lt(x, y) | | minus(x, y) | → | help(lt(y, x), x, y) |
| help(true, x, y) | → | s(minus(x, s(y))) | | help(false, x, y) | → | 0 |
Original Signature
Termination of terms over the following signature is verified: help, 0, minus, s, 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) |