Term Rewriting System R:
[z, x, y]
h(z, e(x)) -> h(c(z), d(z, x))
d(z, g(0, 0)) -> e(0)
d(z, g(x, y)) -> g(e(x), d(z, y))
d(c(z), g(g(x, y), 0)) -> g(d(c(z), g(x, y)), d(z, g(x, y)))
g(e(x), e(y)) -> e(g(x, y))
Termination of R to be shown.
R
↳Dependency Pair Analysis
R contains the following Dependency Pairs:
H(z, e(x)) -> H(c(z), d(z, x))
H(z, e(x)) -> D(z, x)
D(z, g(x, y)) -> G(e(x), d(z, y))
D(z, g(x, y)) -> D(z, y)
D(c(z), g(g(x, y), 0)) -> G(d(c(z), g(x, y)), d(z, g(x, y)))
D(c(z), g(g(x, y), 0)) -> D(c(z), g(x, y))
D(c(z), g(g(x, y), 0)) -> D(z, g(x, y))
G(e(x), e(y)) -> G(x, y)
Furthermore, R contains three SCCs.
R
↳DPs
→DP Problem 1
↳Size-Change Principle
→DP Problem 2
↳MRR
→DP Problem 3
↳Inst
Dependency Pair:
G(e(x), e(y)) -> G(x, y)
Rules:
h(z, e(x)) -> h(c(z), d(z, x))
d(z, g(0, 0)) -> e(0)
d(z, g(x, y)) -> g(e(x), d(z, y))
d(c(z), g(g(x, y), 0)) -> g(d(c(z), g(x, y)), d(z, g(x, y)))
g(e(x), e(y)) -> e(g(x, y))
We number the DPs as follows:
- G(e(x), e(y)) -> G(x, y)
and get the following Size-Change Graph(s):
which lead(s) to this/these maximal multigraph(s):
DP: empty set
Oriented Rules: none
We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial
with Argument Filtering System:
e(x1) -> e(x1)
We obtain no new DP problems.
R
↳DPs
→DP Problem 1
↳SCP
→DP Problem 2
↳Modular Removal of Rules
→DP Problem 3
↳Inst
Dependency Pairs:
D(c(z), g(g(x, y), 0)) -> D(z, g(x, y))
D(c(z), g(g(x, y), 0)) -> D(c(z), g(x, y))
D(z, g(x, y)) -> D(z, y)
Rules:
h(z, e(x)) -> h(c(z), d(z, x))
d(z, g(0, 0)) -> e(0)
d(z, g(x, y)) -> g(e(x), d(z, y))
d(c(z), g(g(x, y), 0)) -> g(d(c(z), g(x, y)), d(z, g(x, y)))
g(e(x), e(y)) -> e(g(x, y))
We have the following set of usable rules:
g(e(x), e(y)) -> e(g(x, y))
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
POL(c(x1)) | = 1 + x1 |
POL(0) | = 0 |
POL(g(x1, x2)) | = x1 + x2 |
POL(e(x1)) | = x1 |
POL(D(x1, x2)) | = 1 + x1 + x2 |
We have the following set D of usable symbols: {c, g, e, D}
The following Dependency Pairs can be deleted as they contain symbols in their lhs which do not occur in D:
D(c(z), g(g(x, y), 0)) -> D(z, g(x, y))
D(c(z), g(g(x, y), 0)) -> D(c(z), g(x, y))
4 non usable rules have been deleted.
The result of this processor delivers one new DP problem.
R
↳DPs
→DP Problem 1
↳SCP
→DP Problem 2
↳MRR
→DP Problem 4
↳Modular Removal of Rules
→DP Problem 3
↳Inst
Dependency Pair:
D(z, g(x, y)) -> D(z, y)
Rule:
g(e(x), e(y)) -> e(g(x, y))
We have the following set of usable rules:
none
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
POL(g(x1, x2)) | = x1 + x2 |
POL(D(x1, x2)) | = x1 + x2 |
We have the following set D of usable symbols: {D}
The following Dependency Pairs can be deleted as they contain symbols in their lhs which do not occur in D:
D(z, g(x, y)) -> D(z, y)
No Rules can be deleted.
After the removal, there are no SCCs in the dependency graph which results in no DP problems which have to be solved.
R
↳DPs
→DP Problem 1
↳SCP
→DP Problem 2
↳MRR
→DP Problem 3
↳Instantiation Transformation
Dependency Pair:
H(z, e(x)) -> H(c(z), d(z, x))
Rules:
h(z, e(x)) -> h(c(z), d(z, x))
d(z, g(0, 0)) -> e(0)
d(z, g(x, y)) -> g(e(x), d(z, y))
d(c(z), g(g(x, y), 0)) -> g(d(c(z), g(x, y)), d(z, g(x, y)))
g(e(x), e(y)) -> e(g(x, y))
On this DP problem, an Instantiation SCC transformation can be performed.
As a result of transforming the rule
H(z, e(x)) -> H(c(z), d(z, x))
one new Dependency Pair
is created:
H(c(z''), e(x')) -> H(c(c(z'')), d(c(z''), x'))
The transformation is resulting in one new DP problem:
R
↳DPs
→DP Problem 1
↳SCP
→DP Problem 2
↳MRR
→DP Problem 3
↳Inst
→DP Problem 5
↳Remaining Obligation(s)
The following remains to be proven:
Dependency Pair:
H(c(z''), e(x')) -> H(c(c(z'')), d(c(z''), x'))
Rules:
h(z, e(x)) -> h(c(z), d(z, x))
d(z, g(0, 0)) -> e(0)
d(z, g(x, y)) -> g(e(x), d(z, y))
d(c(z), g(g(x, y), 0)) -> g(d(c(z), g(x, y)), d(z, g(x, y)))
g(e(x), e(y)) -> e(g(x, y))
The Proof could not be continued due to a Timeout.
Termination of R could not be shown.
Duration:
1:00 minutes