Certification Problem

Input (COPS 577)

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)
+(0,y) y (6)
+(p(x),y) p(+(x,y)) (7)
+(s(x),y) s(+(x,y)) (8)

The underlying signature is as follows:

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

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

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:

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

1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[s(x1)] = 1 · x1 + 6
[+(x1, x2)] = 1 · x1 + 1 · x2 + 1
[p(x1)] = 1 · x1 + 3
[0] = 0
all of the following rules can be deleted.
s(p(x)) x (1)
p(s(x)) x (2)
+(x,0) x (3)
+(0,y) y (6)

1.1.1.1 Rule Removal

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

1.1.1.1.1 Rule Removal

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

1.1.1.1.1.1 Rule Removal

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

1.1.1.1.1.1.1 R is empty

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