YES Confluence Proof

Confluence Proof

by csi

Input

The rewrite relation of the following TRS is considered.

f(g(x),g(y)) → f(g(x),h(y))
f(h(x),g(y)) → f(g(x),g(y))
f(g(x),h(y)) → f(x,y)
f(h(x),h(y)) → f(y,x)
f(x,y) → f(y,x)
g(x) → h(x)
h(x) → g(x)

Proof

1 Redundant Rules Transformation

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

h(x) → g(x)
g(x) → h(x)
f(x,y) → f(y,x)
f(h(x),h(y)) → f(y,x)
f(g(x),h(y)) → f(x,y)

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(x) → g(x)
g(x) → h(x)
f(x,y) → f(y,x)
f(h(x),h(y)) → f(y,x)
f(g(x),h(y)) → f(x,y)
f(h(y),g(x)) → f(x,y)
f(g(x),g(y)) → f(x,y)
f(h(x),g(y)) → f(y,x)
f(h(x85),g(x84)) → f(x84,x85)

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

1.1.1 Strongly closed

Confluence is proven since the TRS is strongly closed. The joins can be performed within 7 step(s).

Tool configuration

csi