Certification Problem

Input (COPS 581)

We consider the TRS containing the following rules:

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

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:

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

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:

+(y,s(0)) s(+(0,y)) (6)
+(y,0) y (5)
+(0,y) y (1)
+(s(0),y) s(+(0,y)) (2)

1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[+(x1, x2)] = 1 · x1 + 1 · x2 + 0
[0] = 4
[s(x1)] = 1 · x1 + 4
all of the following rules can be deleted.
+(y,0) y (5)
+(0,y) y (1)

1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[+(x1, x2)] = 2 · x1 + 2 · x2 + 1
[0] = 7
[s(x1)] = 1 · x1 + 1
all of the following rules can be deleted.
+(y,s(0)) s(+(0,y)) (6)
+(s(0),y) s(+(0,y)) (2)

1.1.1.1.1 R is empty

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