Certification Problem
Input (COPS 50)
We consider the TRS containing the following 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) |
The underlying signature is as follows:
{F/1, G/3, A/0, B/0, H/2, C/0, D/0, K1/1, K2/1}Property / Task
Prove or disprove confluence.Answer / Result
Yes.Proof (by csi @ CoCo 2021)
1 Redundant Rules Transformation
To prove that the TRS is (non-)confluent, we show (non-)confluence of the following
modified system:
|
K2(B) |
→ |
D |
(4) |
|
K1(A) |
→ |
C |
(3) |
|
G(F(H(C,D)),x,y) |
→ |
H(K1(x),K2(y)) |
(2) |
|
F(G(x,A,B)) |
→ |
x |
(1) |
All redundant rules that were added or removed can be
simulated in 2 steps
.
1.1 Critical Pair Closing System
Confluence is proven using the following terminating critical-pair-closing-system R:
|
K1(A) |
→ |
C |
(3) |
|
K2(B) |
→ |
D |
(4) |
1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [C] |
= |
0 |
| [K2(x1)] |
= |
4 · x1 + 1 |
| [B] |
= |
0 |
| [D] |
= |
0 |
| [A] |
= |
0 |
| [K1(x1)] |
= |
2 · x1 + 0 |
all of the following rules can be deleted.
1.1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [C] |
= |
0 |
| [A] |
= |
4 |
| [K1(x1)] |
= |
1 · x1 + 1 |
all of the following rules can be deleted.
1.1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.