YES Confluence Proof

Confluence Proof

by csi

Input

The rewrite relation of the following TRS is considered.

F(H(x),y) F(H(x),I(I(y)))
F(x,G(y)) F(I(x),G(y))
I(x) x

Proof

1 Redundant Rules Transformation

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

I(x) x
F(x,G(y)) F(I(x),G(y))
F(H(x),y) F(H(x),I(I(y)))
F(H(x29),I(I(G(y)))) F(H(x29),G(y))
F(I(H(x)),G(x28)) F(H(x),I(I(G(x28))))
F(I(H(x29)),G(y)) F(H(x29),I(I(G(y))))
F(H(x),I(I(G(x28)))) F(H(x),G(x28))

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

1.1 Development Closed

Confluence is proven since the TRS is development closed.

Tool configuration

csi