Certification Problem

Input (COPS 146)

We consider the TRS containing the following rules:

+(x,0) x (1)
+(s(x),y) s(+(y,x)) (2)
+(x,y) +(y,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:

+(x,y) +(y,x) (3)
+(s(x),y) s(+(y,x)) (2)
+(x,0) x (1)
s(+(0,x33)) s(x33) (4)
+(0,x) x (5)
+(y,s(x)) s(+(y,x)) (6)
s(+(0,x)) s(x) (7)
+(y,s(x31)) s(+(y,x31)) (8)

All redundant rules that were added or removed can be simulated in 2 steps .

1.1 Strongly closed

Confluence is proven since the TRS is strongly closed. The joins can be performed using 7 step(s).