Certification Problem
Input (COPS 1069)
We consider two TRSs R and S where R contains the rules
|
+(0,0) |
→ |
0 |
(1) |
|
+(s(x),y) |
→ |
s(+(x,y)) |
(2) |
|
+(x,s(y)) |
→ |
s(+(y,x)) |
(3) |
and S contains the following rules:
|
+(0,y) |
→ |
y |
(4) |
|
+(s(x),y) |
→ |
s(+(y,x)) |
(5) |
|
+(x,y) |
→ |
+(y,x) |
(6) |
The underlying signature is as follows:
{+/2, 0/0, s/1}Property / Task
Prove or disprove commutation.Answer / Result
Yes.Proof (by ACP @ CoCo 2023)
1 Swap TRSs
The role of TRSs R and S are changed.
1.1 Development Closed
Commutation is proven since the TRSs are development closed.
The joins can be performed using
5
step(s).