YES Confluence Proof

Confluence Proof

by csi

Input

The rewrite relation of the following TRS is considered.

s(p(x)) → x
p(s(x)) → x
+(x,0) → x
+(x,s(y)) → s(+(x,y))
+(x,p(y)) → p(+(x,y))
+(0,y) → y
+(p(x),y) → p(+(x,y))
+(s(x),y) → s(+(x,y))

Proof

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))
+(p(x),y) → p(+(x,y))
+(0,y) → y
+(x,p(y)) → p(+(x,y))
+(x,s(y)) → s(+(x,y))
+(x,0) → x
p(s(x)) → x
s(p(x)) → x

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))
s(p(x)) → x
+(s(x),y) → s(+(x,y))
p(s(x)) → x
+(x,s(y)) → s(+(x,y))
+(x,0) → x
+(p(x),y) → p(+(x,y))
+(0,y) → y

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
the rules
+(x,p(y)) → p(+(x,y))
+(s(x),y) → s(+(x,y))
+(x,s(y)) → s(+(x,y))
+(p(x),y) → p(+(x,y))
remain.

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
the rules
+(s(x),y) → s(+(x,y))
+(x,s(y)) → s(+(x,y))
+(p(x),y) → p(+(x,y))
remain.

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
the rules
+(s(x),y) → s(+(x,y))
+(x,s(y)) → s(+(x,y))
remain.

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 rules could be removed.

1.1.1.1.1.1.1 R is empty

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

Tool configuration

csi