Certification Problem
Input (COPS 99)
We consider the TRS containing the following rules:
|
W(W(x)) |
→ |
W(x) |
(1) |
|
B(I(x)) |
→ |
W(x) |
(2) |
|
W(B(x)) |
→ |
B(x) |
(3) |
|
F(H(x),y) |
→ |
F(H(x),G(y)) |
(4) |
|
F(x,I(y)) |
→ |
F(G(x),I(y)) |
(5) |
|
G(x) |
→ |
x |
(6) |
The underlying signature is as follows:
{W/1, B/1, I/1, F/2, H/1, G/1}Property / Task
Prove or disprove confluence.Answer / Result
Yes.Proof (by csi @ CoCo 2020)
1 Redundant Rules Transformation
To prove that the TRS is (non-)confluent, we show (non-)confluence of the following
modified system:
|
W(W(x)) |
→ |
W(x) |
(1) |
|
B(I(x)) |
→ |
W(x) |
(2) |
|
W(B(x)) |
→ |
B(x) |
(3) |
|
G(x) |
→ |
x |
(6) |
All redundant rules that were added or removed can be
simulated in 4 steps
.
1.1 Critical Pair Closing System
Confluence is proven using the following terminating critical-pair-closing-system R:
|
W(W(x)) |
→ |
W(x) |
(1) |
|
B(I(x)) |
→ |
W(x) |
(2) |
1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [B(x1)] |
= |
1 · x1 + 1 |
| [I(x1)] |
= |
4 · x1 + 4 |
| [W(x1)] |
= |
2 · x1 + 5 |
all of the following rules can be deleted.
1.1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [B(x1)] |
= |
4 · x1 + 1 |
| [I(x1)] |
= |
1 · x1 + 2 |
| [W(x1)] |
= |
4 · x1 + 0 |
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.