Certification Problem

Input (COPS 585)

We consider the TRS containing the following rules:

+(0,y) → y (1)
+(s(x),y) → s(+(x,y)) (2)
s(s(x)) → x (3)

The underlying signature is as follows:

{+/2, 0/0, s/1}

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:

s(s(x)) → x (3)
+(s(x),y) → s(+(x,y)) (2)
+(0,y) → y (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:

+(s(x),y) → s(+(x,y)) (2)
s(s(x)) → x (3)

1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[+(x1, x2)] = 6 · x1 + 1 · x2 + 2
[s(x1)] = 1 · x1 + 2
all of the following rules can be deleted.
+(s(x),y) → s(+(x,y)) (2)
s(s(x)) → x (3)

1.1.1.1 R is empty

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