Termination Proof Script
Consider the TRS R consisting of the rewrite rules
|
1: |
|
x : x |
→ e |
2: |
|
x : e |
→ x |
3: |
|
i(x : y) |
→ y : x |
4: |
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(x : y) : z |
→ x : (z : i(y)) |
5: |
|
e : x |
→ i(x) |
6: |
|
i(i(x)) |
→ x |
7: |
|
i(e) |
→ e |
8: |
|
x : (y : i(x)) |
→ i(y) |
9: |
|
x : (y : (i(x) : z)) |
→ i(z) : y |
10: |
|
i(x) : (y : x) |
→ i(y) |
11: |
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i(x) : (y : (x : z)) |
→ i(z) : y |
|
There are 11 dependency pairs:
|
12: |
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I(x : y) |
→ y :# x |
13: |
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(x : y) :# z |
→ x :# (z : i(y)) |
14: |
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(x : y) :# z |
→ z :# i(y) |
15: |
|
(x : y) :# z |
→ I(y) |
16: |
|
e :# x |
→ I(x) |
17: |
|
x :# (y : i(x)) |
→ I(y) |
18: |
|
x :# (y : (i(x) : z)) |
→ i(z) :# y |
19: |
|
x :# (y : (i(x) : z)) |
→ I(z) |
20: |
|
i(x) :# (y : x) |
→ I(y) |
21: |
|
i(x) :# (y : (x : z)) |
→ i(z) :# y |
22: |
|
i(x) :# (y : (x : z)) |
→ I(z) |
|
Consider the SCC {12-22}.
The constraints could not be solved.
Tyrolean Termination Tool (0.04 seconds)
--- May 4, 2006