Certification Problem
Input (COPS 66)
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) |
|
f(h(k1(a),k2(b))) |
→ |
f(h(c,d)) |
(5) |
|
f(h(c,k2(b))) |
→ |
f(h(c,d)) |
(6) |
|
f(h(k1(a),d)) |
→ |
f(h(c,d)) |
(7) |
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 2023)
1 Redundant Rules Transformation
To prove that the TRS is (non-)confluent, we show (non-)confluence of the following
modified system:
|
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) |
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:
|
k2(b) |
→ |
d |
(4) |
|
k1(a) |
→ |
c |
(3) |
1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [c] |
= |
0 |
| [k2(x1)] |
= |
2 · x1 + 0 |
| [b] |
= |
0 |
| [d] |
= |
0 |
| [a] |
= |
0 |
| [k1(x1)] |
= |
4 · x1 + 1 |
all of the following rules can be deleted.
1.1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [k2(x1)] |
= |
1 · x1 + 1 |
| [b] |
= |
4 |
| [d] |
= |
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.