Certification Problem

Input (COPS 158)

We consider the TRS containing the following rules:

+(0,y) y (1)
+(x,0) x (2)
+(s(x),y) s(+(x,y)) (3)
+(x,s(y)) s(+(x,y)) (4)
+(x,y) +(y,x) (5)

The underlying signature is as follows:

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

Property / Task

Prove or disprove confluence.

Answer / Result

Yes.

Proof (by csi @ CoCo 2022)

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

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

1.1 Parallel Closed

Confluence is proven since the TRS is (almost) parallel closed. The joins can be performed using 1 parallel step(s).