YES Confluence Proof

Confluence Proof

by csi

Input

The rewrite relation of the following TRS is considered.

and3(x,y,F) F
and3(T,T,T) T
and3(x,y,z) and3(y,z,x)

Proof

1 Redundant Rules Transformation

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

and3(x,y,z) and3(y,z,x)
and3(T,T,T) T
and3(x,y,F) F
and3(y,F,x) F

All redundant rules that were added or removed can be simulated in 3 steps .

1.1 Redundant Rules Transformation

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

and3(y,F,x) F
and3(x,y,F) F
and3(T,T,T) T
and3(x,y,z) and3(y,z,x)
and3(F,x,y) F
and3(F,z,x) F

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:

and3(F,x,y) F
and3(y,F,x) F
and3(T,T,T) T
and3(x,y,F) F

1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[and3(x1, x2, x3)] = 2 · x1 + 6 · x2 + 1 · x3 + 0
[F] = 5
[T] = 3
the rule
and3(x,y,F) F
remains.

1.1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[and3(x1, x2, x3)] = 2 · x1 + 4 · x2 + 1 · x3 + 1
[F] = 0
all rules could be removed.

1.1.1.1.1.1 R is empty

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

Tool configuration

csi