Certification Problem
Input (COPS 1054)
We consider two TRSs R and S where R contains the rules
|
f(g(x,a,b)) |
→ |
x |
(1) |
|
g(f(h(c,d)),x,y) |
→ |
h(k1(x),k2(y)) |
(2) |
|
k1(a) |
→ |
c |
(3) |
|
k2(b) |
→ |
d |
(4) |
|
f(h(k1(a),k2(b))) |
→ |
f(h(c,d)) |
(5) |
|
f(h(c,k2(b))) |
→ |
f(h(c,d)) |
(6) |
|
f(h(k1(a),d)) |
→ |
f(h(c,d)) |
(7) |
and S contains the following rules:
|
f(a) |
→ |
b |
(8) |
|
f(a) |
→ |
f(c) |
(9) |
| a |
→ |
d |
(10) |
|
f(d) |
→ |
b |
(11) |
|
f(c) |
→ |
b |
(12) |
| d |
→ |
c |
(13) |
The underlying signature is as follows:
{a/0, b/0, c/0, d/0, f/1, g/3, h/2, k1/1, k2/1}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.
and
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.