Certification Problem
Input (COPS 93)
We consider the TRS containing the following rules:
|
F(H(x),y) |
→ |
G(H(x)) |
(1) |
|
H(I(x)) |
→ |
I(x) |
(2) |
|
F(I(x),y) |
→ |
G(I(x)) |
(3) |
The underlying signature is as follows:
{F/2, H/1, G/1, I/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:
|
F(I(x),y) |
→ |
G(I(x)) |
(3) |
|
H(I(x)) |
→ |
I(x) |
(2) |
|
F(H(x),y) |
→ |
G(H(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:
|
F(I(x),y) |
→ |
G(I(x)) |
(3) |
|
H(I(x)) |
→ |
I(x) |
(2) |
1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [I(x1)] |
= |
2 · x1 + 5 |
| [G(x1)] |
= |
2 · x1 + 0 |
| [F(x1, x2)] |
= |
4 · x1 + 4 · x2 + 0 |
| [H(x1)] |
= |
1 · x1 + 0 |
all of the following rules can be deleted.
1.1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [I(x1)] |
= |
1 · x1 + 4 |
| [H(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.