YES Confluence Proof

Confluence Proof

by csi

Input

The rewrite relation of the following TRS is considered.

+(x,0) → x
+(x,s(y)) → s(+(x,y))
d(0) → 0
d(s(x)) → s(s(d(x)))
f(0) → 0
f(s(x)) → +(+(s(x),s(x)),s(x))
f(g(0)) → +(+(g(0),g(0)),g(0))
g(x) → s(d(x))

Proof

1 Redundant Rules Transformation

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

+(x,0) → x
+(x,s(y)) → s(+(x,y))
d(0) → 0
d(s(x)) → s(s(d(x)))
f(0) → 0
f(s(x)) → +(+(s(x),s(x)),s(x))
g(x) → s(d(x))

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:

There are no rules.

1.1.1 R is empty

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

Tool configuration

csi