TIMEOUT
The TRS could not be proven terminating. The proof attempt took 60012 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (49ms).
| – Problem 2 was processed with processor SubtermCriterion (1ms).
| – Problem 3 was processed with processor BackwardInstantiation (5ms).
| | – Problem 5 remains open; application of the following processors failed [ForwardInstantiation (1ms), Propagation (0ms), ForwardNarrowing (0ms), BackwardInstantiation (1ms), ForwardInstantiation (3ms), Propagation (1ms)].
| – Problem 4 was processed with processor SubtermCriterion (1ms).
The following open problems remain:
Open Dependency Pair Problem 3
Dependency Pairs
| if#(false, x, s(y), z) | → | quot#(minus(x, s(y)), s(y), z) | | quot#(x, y, z) | → | if#(zero(x), x, y, plus(z, s(0))) |
Rewrite Rules
| minus(0, y) | → | 0 | | minus(x, 0) | → | x |
| minus(s(x), s(y)) | → | minus(x, y) | | plus(0, y) | → | y |
| plus(s(x), y) | → | plus(x, s(y)) | | zero(s(x)) | → | false |
| zero(0) | → | true | | p(s(x)) | → | x |
| div(x, y) | → | quot(x, y, 0) | | quot(x, y, z) | → | if(zero(x), x, y, plus(z, s(0))) |
| if(true, x, y, z) | → | p(z) | | if(false, x, s(y), z) | → | quot(minus(x, s(y)), s(y), z) |
Original Signature
Termination of terms over the following signature is verified: plus, 0, minus, s, if, p, div, true, false, zero, quot
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| div#(x, y) | → | quot#(x, y, 0) | | quot#(x, y, z) | → | zero#(x) |
| if#(true, x, y, z) | → | p#(z) | | if#(false, x, s(y), z) | → | minus#(x, s(y)) |
| minus#(s(x), s(y)) | → | minus#(x, y) | | if#(false, x, s(y), z) | → | quot#(minus(x, s(y)), s(y), z) |
| plus#(s(x), y) | → | plus#(x, s(y)) | | quot#(x, y, z) | → | plus#(z, s(0)) |
| quot#(x, y, z) | → | if#(zero(x), x, y, plus(z, s(0))) |
Rewrite Rules
| minus(0, y) | → | 0 | | minus(x, 0) | → | x |
| minus(s(x), s(y)) | → | minus(x, y) | | plus(0, y) | → | y |
| plus(s(x), y) | → | plus(x, s(y)) | | zero(s(x)) | → | false |
| zero(0) | → | true | | p(s(x)) | → | x |
| div(x, y) | → | quot(x, y, 0) | | quot(x, y, z) | → | if(zero(x), x, y, plus(z, s(0))) |
| if(true, x, y, z) | → | p(z) | | if(false, x, s(y), z) | → | quot(minus(x, s(y)), s(y), z) |
Original Signature
Termination of terms over the following signature is verified: plus, minus, 0, s, if, p, div, false, true, zero, quot
Strategy
The following SCCs where found
| if#(false, x, s(y), z) → quot#(minus(x, s(y)), s(y), z) | quot#(x, y, z) → if#(zero(x), x, y, plus(z, s(0))) |
| minus#(s(x), s(y)) → minus#(x, y) |
| plus#(s(x), y) → plus#(x, s(y)) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| minus#(s(x), s(y)) | → | minus#(x, y) |
Rewrite Rules
| minus(0, y) | → | 0 | | minus(x, 0) | → | x |
| minus(s(x), s(y)) | → | minus(x, y) | | plus(0, y) | → | y |
| plus(s(x), y) | → | plus(x, s(y)) | | zero(s(x)) | → | false |
| zero(0) | → | true | | p(s(x)) | → | x |
| div(x, y) | → | quot(x, y, 0) | | quot(x, y, z) | → | if(zero(x), x, y, plus(z, s(0))) |
| if(true, x, y, z) | → | p(z) | | if(false, x, s(y), z) | → | quot(minus(x, s(y)), s(y), z) |
Original Signature
Termination of terms over the following signature is verified: plus, minus, 0, s, if, p, div, false, true, zero, quot
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| minus#(s(x), s(y)) | → | minus#(x, y) |
Problem 3: BackwardInstantiation
Dependency Pair Problem
Dependency Pairs
| if#(false, x, s(y), z) | → | quot#(minus(x, s(y)), s(y), z) | | quot#(x, y, z) | → | if#(zero(x), x, y, plus(z, s(0))) |
Rewrite Rules
| minus(0, y) | → | 0 | | minus(x, 0) | → | x |
| minus(s(x), s(y)) | → | minus(x, y) | | plus(0, y) | → | y |
| plus(s(x), y) | → | plus(x, s(y)) | | zero(s(x)) | → | false |
| zero(0) | → | true | | p(s(x)) | → | x |
| div(x, y) | → | quot(x, y, 0) | | quot(x, y, z) | → | if(zero(x), x, y, plus(z, s(0))) |
| if(true, x, y, z) | → | p(z) | | if(false, x, s(y), z) | → | quot(minus(x, s(y)), s(y), z) |
Original Signature
Termination of terms over the following signature is verified: plus, minus, 0, s, if, p, div, false, true, zero, quot
Strategy
Instantiation
For all potential predecessors l → r of the rule quot
#(
x,
y,
z) → if
#(zero(
x),
x,
y, plus(
z, s(0))) on dependency pair chains it holds that:
- quot#(x, y, z) matches r,
- all variables of quot#(x, y, z) are embedded in constructor contexts, i.e., each subterm of quot#(x, y, z), containing a variable is rooted by a constructor symbol.
Thus, quot
#(
x,
y,
z) → if
#(zero(
x),
x,
y, plus(
z, s(0))) is replaced by instances determined through the above matching. These instances are:
| quot#(minus(_x, s(_y)), s(_y), _z) → if#(zero(minus(_x, s(_y))), minus(_x, s(_y)), s(_y), plus(_z, s(0))) |
Problem 4: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| plus#(s(x), y) | → | plus#(x, s(y)) |
Rewrite Rules
| minus(0, y) | → | 0 | | minus(x, 0) | → | x |
| minus(s(x), s(y)) | → | minus(x, y) | | plus(0, y) | → | y |
| plus(s(x), y) | → | plus(x, s(y)) | | zero(s(x)) | → | false |
| zero(0) | → | true | | p(s(x)) | → | x |
| div(x, y) | → | quot(x, y, 0) | | quot(x, y, z) | → | if(zero(x), x, y, plus(z, s(0))) |
| if(true, x, y, z) | → | p(z) | | if(false, x, s(y), z) | → | quot(minus(x, s(y)), s(y), z) |
Original Signature
Termination of terms over the following signature is verified: plus, minus, 0, s, if, p, div, false, true, zero, quot
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| plus#(s(x), y) | → | plus#(x, s(y)) |