Certification Problem

Input (COPS 576)

We consider the TRS containing the following rules:

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

The underlying signature is as follows:

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

Property / Task

Prove or disprove confluence.

Answer / Result

Yes.

Proof (by csi @ CoCo 2022)

1 Critical Pair Closing System

Confluence is proven using the following terminating critical-pair-closing-system R:

+(x,p(y)) → p(+(x,y)) (5)
s(p(x)) → x (1)
-(x,p(y)) → s(-(x,y)) (8)
p(s(x)) → x (2)
+(x,s(y)) → s(+(x,y)) (4)
-(x,s(y)) → p(-(x,y)) (7)

1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[-(x1, x2)] = 1 · x1 + 6 · x2 + 0
[s(x1)] = 1 · x1 + 1
[+(x1, x2)] = 4 · x1 + 2 · x2 + 0
[p(x1)] = 1 · x1 + 4
all of the following rules can be deleted.
+(x,p(y)) → p(+(x,y)) (5)
s(p(x)) → x (1)
-(x,p(y)) → s(-(x,y)) (8)
p(s(x)) → x (2)
+(x,s(y)) → s(+(x,y)) (4)
-(x,s(y)) → p(-(x,y)) (7)

1.1.1 R is empty

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