Certification Problem

Input (COPS 33)

We consider the TRS containing the following rules:

f(a) f(g(b,b)) (1)
a g(c,c) (2)
c d (3)
d b (4)
b d (5)

The underlying signature is as follows:

{f/1, a/0, g/2, b/0, c/0, d/0}

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:

b d (5)
d b (4)
c d (3)
a g(c,c) (2)
f(a) f(g(b,b)) (1)
f(g(c,c)) f(g(d,d)) (6)
f(g(b,b)) f(g(d,d)) (7)

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:

b d (5)
c d (3)
f(g(c,c)) f(g(d,d)) (6)
f(g(b,b)) f(g(d,d)) (7)

1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[d] = 2
[c] = 4
[g(x1, x2)] = 2 · x1 + 4 · x2 + 1
[b] = 4
[f(x1)] = 1 · x1 + 2
all of the following rules can be deleted.
b d (5)
c d (3)
f(g(c,c)) f(g(d,d)) (6)
f(g(b,b)) f(g(d,d)) (7)

1.1.1.1 R is empty

There are no rules in the TRS. Hence, it is terminating.