Certification Problem

Input (COPS 84)

We consider the TRS containing the following rules:

a b (1)
f(a) g(a) (2)
f(b) g(b) (3)

The underlying signature is as follows:

{a/0, b/0, f/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:

f(b) g(b) (3)
f(a) g(a) (2)
a b (1)
f(a) g(b) (4)

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:

a b (1)
f(b) g(b) (3)

1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[b] = 0
[g(x1)] = 1 · x1 + 0
[a] = 0
[f(x1)] = 1 · x1 + 1
all of the following rules can be deleted.
f(b) g(b) (3)

1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[b] = 0
[a] = 1
all of the following rules can be deleted.
a b (1)

1.1.1.1.1 R is empty

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