Certification Problem
Input (COPS 1075)
We consider two TRSs R and S where R contains the rules
|
f(g(f(x))) |
→ |
g(f(g(x))) |
(1) |
|
f(c) |
→ |
c |
(2) |
|
g(c) |
→ |
c |
(3) |
and S contains the following rules:
|
f(a) |
→ |
b |
(4) |
| a |
→ |
a' |
(5) |
|
f(b) |
→ |
c |
(6) |
The underlying signature is as follows:
{a/0, b/0, c/0, f/1, g/1, a'/0}Property / Task
Prove or disprove commutation.Answer / Result
No.Proof (by ACP @ CoCo 2023)
1 Non-Joinable Fork
The systems are not commuting due to the following forking derivations.
| t0
|
= |
f(g(f(a))) |
|
→S
|
f(g(b)) |
|
= |
s1
|
and
| t0
|
= |
f(g(f(a))) |
|
→R
|
g(f(g(a))) |
|
= |
t1
|
There is no possibility to join
s1→R*·←S*
t1
for the following reason:
- When applying the cap-function on both terms (where variables may be treated like constants)
then the resulting terms do not unify.