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 2021)

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
[F(x1, x2)] = 5 · x1 + 4 · x2 + 6
[I(x1)] = 1 · x1 + 2
[H(x1)] = 1 · x1 + 1
[G(x1)] = 5 · x1 + 2
all of the following rules can be deleted.
F(I(x),y) → G(I(x)) (3)
H(I(x)) → I(x) (2)

1.1.1.1 R is empty

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