YES Confluence Proof

Confluence Proof

by csi

Input

The rewrite relation of the following TRS is considered.

h(f(f(c)),b) f(h(h(h(c,h(f(h(c,f(b))),a)),b),c))
c c
f(f(h(h(f(a),a),c))) f(h(f(c),b))
h(f(h(f(b),h(h(f(h(c,f(c))),b),a))),h(a,c)) c

Proof

1 Redundant Rules Transformation

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

h(f(h(f(b),h(h(f(h(c,f(c))),b),a))),h(a,c)) c
f(f(h(h(f(a),a),c))) f(h(f(c),b))
h(f(f(c)),b) f(h(h(h(c,h(f(h(c,f(b))),a)),b),c))

All redundant rules that were added or removed can be simulated in 1 steps .

1.1 Redundant Rules Transformation

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

h(f(h(f(b),h(h(f(h(c,f(c))),b),a))),h(a,c)) c
f(f(h(h(f(a),a),c))) f(h(f(c),b))
h(f(f(c)),b) f(h(h(h(c,h(f(h(c,f(b))),a)),b),c))

All redundant rules that were added or removed can be simulated in 2 steps .

1.1.1 Critical Pair Closing System

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

There are no rules.

1.1.1.1 R is empty

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

Tool configuration

csi