Certification Problem

Input (COPS 67)

We consider the TRS containing the following rules:

f(i(x),g(a)) → f(j(x,x),g(b)) (1)
b → a (2)
i(x) → j(x,x) (3)

The underlying signature is as follows:

{f/2, i/1, g/1, a/0, j/2, b/0}

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:

i(x) → j(x,x) (3)
b → a (2)
f(i(x),g(a)) → f(j(x,x),g(b)) (1)
f(i(x),g(a)) → f(j(x,x),g(a)) (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:

b → a (2)

1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[a] = 0
[b] = 1
all of the following rules can be deleted.
b → a (2)

1.1.1.1 R is empty

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