Certification Problem

Input (COPS 63)

We consider the TRS containing the following rules:

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

The underlying signature is as follows:

{-/2, 0/0, s/1, d/1}

Property / Task

Prove or disprove confluence.

Answer / Result

Yes.

Proof (by csi @ CoCo 2020)

1 Redundant Rules Transformation

To prove that the TRS is (non-)confluent, we show (non-)confluence of the following modified system:

-(0,0) 0 (1)
-(s(x),0) s(x) (2)
d(s(x)) x (4)
-(s(x),s(y)) -(x,y) (5)
-(d(x),y) -(x,s(y)) (6)

All redundant rules that were added or removed can be simulated in 4 steps .

1.1 Critical Pair Closing System

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

-(s(x),s(y)) -(x,y) (5)

1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[-(x1, x2)] = 3 · x1 + 6 · x2 + 0
[s(x1)] = 1 · x1 + 3
all of the following rules can be deleted.
-(s(x),s(y)) -(x,y) (5)

1.1.1.1 R is empty

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