Certification Problem
Input (COPS 765)
We consider the TRS containing the following rules:
|
xor(x,0) |
→ |
x |
(1) |
|
xor(0,x) |
→ |
x |
(2) |
|
xor(x,x) |
→ |
0 |
(3) |
|
and(x,0) |
→ |
0 |
(4) |
|
and(0,x) |
→ |
0 |
(5) |
|
and(x,1) |
→ |
x |
(6) |
|
and(1,x) |
→ |
x |
(7) |
|
and(x,x) |
→ |
x |
(8) |
|
xor(xor(x,y),z) |
→ |
xor(x,xor(y,z)) |
(9) |
|
xor(x,xor(y,z)) |
→ |
xor(y,xor(x,z)) |
(10) |
|
xor(x,y) |
→ |
xor(y,x) |
(11) |
|
xor(x,xor(x,y)) |
→ |
y |
(12) |
|
and(and(x,y),z) |
→ |
and(x,and(y,z)) |
(13) |
|
and(x,and(y,z)) |
→ |
and(y,and(x,z)) |
(14) |
|
and(x,y) |
→ |
and(y,x) |
(15) |
|
and(x,and(x,y)) |
→ |
and(x,y) |
(16) |
|
and(x,xor(y,z)) |
→ |
xor(and(x,y),xor(x,z)) |
(17) |
|
and(xor(x,y),z) |
→ |
xor(and(x,z),and(y,z)) |
(18) |
The underlying signature is as follows:
{xor/2, 0/0, and/2, 1/0}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
|
= |
and(0,xor(y,z)) |
|
→
|
0 |
|
= |
t1
|
| t0
|
= |
and(0,xor(y,z)) |
|
→
|
xor(and(0,y),xor(0,z)) |
|
→
|
xor(0,xor(0,z)) |
|
→
|
z |
|
= |
t3
|
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.