Term Rewriting System R:
[y, x]
-(0, y) -> 0
-(x, 0) -> x
-(x, s(y)) -> if(greater(x, s(y)), s(-(x, p(s(y)))), 0)
p(0) -> 0
p(s(x)) -> x
Termination of R to be shown.
R
↳Dependency Pair Analysis
R contains the following Dependency Pairs:
-'(x, s(y)) -> -'(x, p(s(y)))
-'(x, s(y)) -> P(s(y))
Furthermore, R contains one SCC.
R
↳DPs
→DP Problem 1
↳Modular Removal of Rules
Dependency Pair:
-'(x, s(y)) -> -'(x, p(s(y)))
Rules:
-(0, y) -> 0
-(x, 0) -> x
-(x, s(y)) -> if(greater(x, s(y)), s(-(x, p(s(y)))), 0)
p(0) -> 0
p(s(x)) -> x
We have the following set of usable rules:
p(s(x)) -> x
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
| POL(-'(x1, x2)) | = x1 + x2 |
| POL(s(x1)) | = x1 |
| POL(p(x1)) | = x1 |
We have the following set D of usable symbols: {-', s, p}
No Dependency Pairs can be deleted.
4 non usable rules have been deleted.
The result of this processor delivers one new DP problem.
R
↳DPs
→DP Problem 1
↳MRR
→DP Problem 2
↳Modular Removal of Rules
Dependency Pair:
-'(x, s(y)) -> -'(x, p(s(y)))
Rule:
p(s(x)) -> x
We have the following set of usable rules:
p(s(x)) -> x
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
| POL(-'(x1, x2)) | = 1 + x1 + x2 |
| POL(s(x1)) | = 1 + x1 |
| POL(p(x1)) | = x1 |
We have the following set D of usable symbols: {-', s, p}
No Dependency Pairs can be deleted.
The following rules can be deleted as the lhs is strictly greater than the corresponding rhs:
p(s(x)) -> x
The result of this processor delivers one new DP problem.
R
↳DPs
→DP Problem 1
↳MRR
→DP Problem 2
↳MRR
...
→DP Problem 3
↳Dependency Graph
Dependency Pair:
-'(x, s(y)) -> -'(x, p(s(y)))
Rule:
none
Using the Dependency Graph resulted in no new DP problems.
Termination of R successfully shown.
Duration:
0:00 minutes