Certification Problem
Input
We consider two TRSs R and S where R contains the rules
|
f(f(x)) |
→ |
f(g(f(x),f(x))) |
(A) |
| a |
→ |
c |
(C) |
| b |
→ |
d |
(D) |
| e1 |
→ |
d |
(H) |
| e2 |
→ |
d |
(I) |
and S contains the following rules:
|
f(f(x)) |
→ |
f(g(f(x),f(x))) |
(A) |
| a |
→ |
b |
(B) |
| c |
→ |
d |
(E) |
| c |
→ |
e1 |
(F) |
| e1 |
→ |
e2 |
(G) |
| e2 |
→ |
d |
(I) |
The underlying signature is as follows:
{f/1, g/2, a/0, b/0, c/0, d/0, e1/0, e2/0}Property / Task
Prove or disprove commutation.Answer / Result
Yes.Proof (by confluence proof of csi + manual change into commutation proof)
1 Simultaneous Critical Pairs Closed
Commutation is proven by a generalization of Okui's criterion to commutation,
that all simultaneous critical pairs u ⇐R
. →S v
are development closed: u →S* .
⇐R v.
The simultaneous critical pairs can be joined as follows. Here,
↔ is always chosen as an appropriate rewrite relation which
is automatically inferred by the certifier.
-
The critical peak s = f(f(g(f(x11),f(x11))))⇐→ f(g(f(f(x11)),f(f(x11)))) = t can be joined as follows.
s
↔ f(g(f(g(f(x11),f(x11))),f(g(f(x11),f(x11))))) ↔
t
-
The critical peak s = f(g(f(f(x)),f(f(x))))⇐→ f(f(g(f(x),f(x)))) = t can be joined as follows.
s
↔ f(g(f(g(f(x),f(x))),f(g(f(x),f(x))))) ↔
t
-
The critical peak s = f(g(f(f(g(f(x14),f(x14)))),f(f(g(f(x14),f(x14))))))⇐→ f(f(g(f(f(x14)),f(f(x14))))) = t can be joined as follows.
s
↔ f(g(f(g(f(g(f(x14),f(x14))),f(g(f(x14),f(x14))))),f(g(f(g(f(x14),f(x14))),f(g(f(x14),f(x14))))))) ↔
t
-
The critical peak s = c⇐→ b = t can be joined as follows.
s
↔ e1 ↔ e2 ↔ d ↔
t
-
The critical peak s = d⇐→ e2 = t can be joined as follows.
s
↔
t
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