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 (22ms).
| – Problem 2 was processed with processor SubtermCriterion (1ms).
| – Problem 3 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 (1ms), BackwardInstantiation (1ms), ForwardInstantiation (1ms), Propagation (1ms)].
The following open problems remain:
Open Dependency Pair Problem 3
Dependency Pairs
| cond#(true, x, y) | → | minus#(x, s(y)) | | minus#(x, y) | → | cond#(gt(x, y), x, y) |
Rewrite Rules
| minus(x, y) | → | cond(gt(x, y), x, y) | | cond(false, x, y) | → | 0 |
| cond(true, x, y) | → | s(minus(x, s(y))) | | gt(0, v) | → | false |
| gt(s(u), 0) | → | true | | gt(s(u), s(v)) | → | gt(u, v) |
Original Signature
Termination of terms over the following signature is verified: 0, minus, s, true, false, gt, cond
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| cond#(true, x, y) | → | minus#(x, s(y)) | | minus#(x, y) | → | cond#(gt(x, y), x, y) |
| minus#(x, y) | → | gt#(x, y) | | gt#(s(u), s(v)) | → | gt#(u, v) |
Rewrite Rules
| minus(x, y) | → | cond(gt(x, y), x, y) | | cond(false, x, y) | → | 0 |
| cond(true, x, y) | → | s(minus(x, s(y))) | | gt(0, v) | → | false |
| gt(s(u), 0) | → | true | | gt(s(u), s(v)) | → | gt(u, v) |
Original Signature
Termination of terms over the following signature is verified: minus, 0, s, false, true, gt, cond
Strategy
The following SCCs where found
| gt#(s(u), s(v)) → gt#(u, v) |
| cond#(true, x, y) → minus#(x, s(y)) | minus#(x, y) → cond#(gt(x, y), x, y) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| gt#(s(u), s(v)) | → | gt#(u, v) |
Rewrite Rules
| minus(x, y) | → | cond(gt(x, y), x, y) | | cond(false, x, y) | → | 0 |
| cond(true, x, y) | → | s(minus(x, s(y))) | | gt(0, v) | → | false |
| gt(s(u), 0) | → | true | | gt(s(u), s(v)) | → | gt(u, v) |
Original Signature
Termination of terms over the following signature is verified: minus, 0, s, false, true, gt, cond
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| gt#(s(u), s(v)) | → | gt#(u, v) |
Problem 3: BackwardInstantiation
Dependency Pair Problem
Dependency Pairs
| cond#(true, x, y) | → | minus#(x, s(y)) | | minus#(x, y) | → | cond#(gt(x, y), x, y) |
Rewrite Rules
| minus(x, y) | → | cond(gt(x, y), x, y) | | cond(false, x, y) | → | 0 |
| cond(true, x, y) | → | s(minus(x, s(y))) | | gt(0, v) | → | false |
| gt(s(u), 0) | → | true | | gt(s(u), s(v)) | → | gt(u, v) |
Original Signature
Termination of terms over the following signature is verified: minus, 0, s, false, true, gt, cond
Strategy
Instantiation
For all potential predecessors l → r of the rule minus
#(
x,
y) → cond
#(gt(
x,
y),
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) → cond
#(gt(
x,
y),
x,
y) is replaced by instances determined through the above matching. These instances are:
| minus#(_x, s(_y)) → cond#(gt(_x, s(_y)), _x, s(_y)) |
Problem 4: BackwardInstantiation
Dependency Pair Problem
Dependency Pairs
| cond#(true, x, y) | → | minus#(x, s(y)) | | minus#(_x, s(_y)) | → | cond#(gt(_x, s(_y)), _x, s(_y)) |
Rewrite Rules
| minus(x, y) | → | cond(gt(x, y), x, y) | | cond(false, x, y) | → | 0 |
| cond(true, x, y) | → | s(minus(x, s(y))) | | gt(0, v) | → | false |
| gt(s(u), 0) | → | true | | gt(s(u), s(v)) | → | gt(u, v) |
Original Signature
Termination of terms over the following signature is verified: 0, minus, s, true, false, gt, cond
Strategy
Instantiation
For all potential predecessors l → r of the rule minus
#(
_x, s(
_y)) → cond
#(gt(
_x, s(
_y)),
_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)) → cond
#(gt(
_x, s(
_y)),
_x, s(
_y)) is replaced by instances determined through the above matching. These instances are:
| minus#(x, s(y)) → cond#(gt(x, s(y)), x, s(y)) |
Problem 5: Propagation
Dependency Pair Problem
Dependency Pairs
| cond#(true, x, y) | → | minus#(x, s(y)) | | minus#(x, s(y)) | → | cond#(gt(x, s(y)), x, s(y)) |
Rewrite Rules
| minus(x, y) | → | cond(gt(x, y), x, y) | | cond(false, x, y) | → | 0 |
| cond(true, x, y) | → | s(minus(x, s(y))) | | gt(0, v) | → | false |
| gt(s(u), 0) | → | true | | gt(s(u), s(v)) | → | gt(u, v) |
Original Signature
Termination of terms over the following signature is verified: minus, 0, s, false, true, gt, cond
Strategy
The dependency pairs cond
#(true,
x,
y) → minus
#(
x, s(
y)) and minus
#(
x, s(
y)) → cond
#(gt(
x, s(
y)),
x, s(
y)) are consolidated into the rule cond
#(true,
x,
y) → cond
#(gt(
x, s(
y)),
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 cond#(true, x, y) and cond#(gt(x, s(y)), x, s(y))
The dependency pairs cond
#(true,
x,
y) → minus
#(
x, s(
y)) and minus
#(
x, s(
y)) → cond
#(gt(
x, s(
y)),
x, s(
y)) are consolidated into the rule cond
#(true,
x,
y) → cond
#(gt(
x, s(
y)),
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 cond#(true, x, y) and cond#(gt(x, s(y)), x, s(y))
Summary
| Removed Dependency Pairs | Added Dependency Pairs |
|---|
| minus#(x, s(y)) → cond#(gt(x, s(y)), x, s(y)) | cond#(true, x, y) → cond#(gt(x, s(y)), x, s(y)) |
| cond#(true, x, y) → minus#(x, s(y)) | |