Certification Problem
Input (COPS 632)
We consider the TRS containing the following rules:
|
inv(0) |
→ |
0 |
(1) |
|
inv(s(x)) |
→ |
p(inv(x)) |
(2) |
|
inv(p(x)) |
→ |
s(inv(x)) |
(3) |
|
minus(x,0) |
→ |
x |
(4) |
|
minus(x,p(y)) |
→ |
s(minus(x,y)) |
(5) |
|
minus(x,s(y)) |
→ |
p(minus(x,y)) |
(6) |
|
minus(0,x) |
→ |
inv(x) |
(7) |
|
inv(x) |
→ |
minus(0,x) |
(8) |
|
inv(minus(x,y)) |
→ |
minus(y,x) |
(9) |
|
s(p(x)) |
→ |
x |
(10) |
|
p(s(x)) |
→ |
x |
(11) |
The underlying signature is as follows:
{inv/1, 0/0, s/1, p/1, minus/2}Property / Task
Prove or disprove confluence.Answer / Result
No.Proof (by csi @ CoCo 2022)
1 Non-Joinable Fork
The system is not confluent due to the following forking derivations.
| t0
|
= |
inv(minus(x,p(x630))) |
|
→
|
inv(s(minus(x,x630))) |
|
→
|
p(inv(minus(x,x630))) |
|
→
|
p(minus(x630,x)) |
|
= |
t3
|
| t0
|
= |
inv(minus(x,p(x630))) |
|
→
|
minus(p(x630),x) |
|
= |
t1
|
The two resulting terms cannot be joined 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.