Certification Problem

Input (COPS 104)

We consider the TRS containing the following rules:

Ap(Ap(Ap(S,x),y),z) → Ap(Ap(x,z),Ap(y,z)) (1)
Ap(Ap(K,x),y) → x (2)
Ap(I,x) → x (3)

The underlying signature is as follows:

{Ap/2, S/0, K/0, I/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:

Ap(I,x) → x (3)
Ap(Ap(K,x),y) → x (2)
Ap(Ap(Ap(S,x),y),z) → Ap(Ap(x,z),Ap(y,z)) (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:

There are no rules.

1.1.1 R is empty

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