Certification Problem

Input (COPS 130)

We consider the TRS containing the following rules:

and3(x,y,F) → F (1)
and3(T,T,T) → T (2)
and3(x,y,z) → and3(y,z,x) (3)

The underlying signature is as follows:

{and3/3, F/0, T/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:

and3(x,y,z) → and3(y,z,x) (3)
and3(T,T,T) → T (2)
and3(x,y,F) → F (1)
and3(y,F,x) → F (4)

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 (4)
and3(x,y,F) → F (1)
and3(T,T,T) → T (2)
and3(x,y,z) → and3(y,z,x) (3)
and3(F,x,y) → F (5)
and3(F,z,x) → F (6)

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 (5)
and3(y,F,x) → F (4)
and3(T,T,T) → T (2)
and3(x,y,F) → F (1)

1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[T] = 1
[and3(x1, x2, x3)] = 4 · x1 + 2 · x2 + 1 · x3 + 0
[F] = 0
all of the following rules can be deleted.
and3(T,T,T) → T (2)

1.1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[and3(x1, x2, x3)] = 4 · x1 + 1 · x2 + 4 · x3 + 0
[F] = 2
all of the following rules can be deleted.
and3(F,x,y) → F (5)
and3(x,y,F) → F (1)

1.1.1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[and3(x1, x2, x3)] = 4 · x1 + 1 · x2 + 2 · x3 + 1
[F] = 0
all of the following rules can be deleted.
and3(y,F,x) → F (4)

1.1.1.1.1.1.1 R is empty

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