Certification Problem

Input (COPS 190)

We consider the TRS containing the following rules:

or(x,T) T (1)
or(x,F) x (2)
or(x,y) or(y,x) (3)

The underlying signature is as follows:

{or/2, T/0, F/0}

Property / Task

Prove or disprove confluence.

Answer / Result

Yes.

Proof (by csi @ CoCo 2020)

1 Redundant Rules Transformation

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

or(x,y) or(y,x) (3)
or(x,F) x (2)
or(x,T) T (1)
or(T,x) T (4)
or(F,x) x (5)

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

1.1 Critical Pair Closing System

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

or(F,x) x (5)
or(T,x) T (4)
or(x,T) T (1)
or(x,F) x (2)

1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[F] = 1
[or(x1, x2)] = 1 · x1 + 4 · x2 + 0
[T] = 5
all of the following rules can be deleted.
or(F,x) x (5)
or(x,T) T (1)
or(x,F) x (2)

1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[or(x1, x2)] = 1 · x1 + 4 · x2 + 1
[T] = 0
all of the following rules can be deleted.
or(T,x) T (4)

1.1.1.1.1 R is empty

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