Certification Problem

Input (COPS 944)

We consider the TRS containing the following rules:

C(x) c(x) (1)
c(c(x)) x (2)
b(b(x)) B(x) (3)
B(B(x)) b(x) (4)
c(B(c(b(c(x))))) B(c(b(c(B(c(b(x))))))) (5)
b(B(x)) x (6)
B(b(x)) x (7)
c(C(x)) x (8)
C(c(x)) x (9)

The underlying signature is as follows:

{C/1, c/1, b/1, B/1}

Property / Task

Prove or disprove confluence.

Answer / Result

Yes.

Proof (by csi @ CoCo 2022)

1 Locally confluent and terminating

Confluence is proven by showing local confluence and termination.

1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[c(x1)] = 2 · x1 + 1
[B(x1)] = 1 · x1 + 0
[C(x1)] = 2 · x1 + 1
[b(x1)] = 1 · x1 + 0
all of the following rules can be deleted.
c(c(x)) x (2)
c(C(x)) x (8)
C(c(x)) x (9)

1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[c(x1)] = 1 · x1 + 0
[B(x1)] = 1 · x1 + 0
[C(x1)] = 2 · x1 + 4
[b(x1)] = 1 · x1 + 0
all of the following rules can be deleted.
C(x) c(x) (1)

1.1.1.1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS
b(b(x)) B(x) (3)
B(B(x)) b(x) (4)
c(b(c(B(c(x))))) b(c(B(c(b(c(B(x))))))) (10)
B(b(x)) x (7)
b(B(x)) x (6)

1.1.1.1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
b#(b(x)) B#(x) (11)
B#(B(x)) b#(x) (12)
c#(b(c(B(c(x))))) B#(x) (13)
c#(b(c(B(c(x))))) c#(B(x)) (14)
c#(b(c(B(c(x))))) b#(c(B(x))) (15)
c#(b(c(B(c(x))))) c#(b(c(B(x)))) (16)
c#(b(c(B(c(x))))) B#(c(b(c(B(x))))) (17)
c#(b(c(B(c(x))))) c#(B(c(b(c(B(x)))))) (18)
c#(b(c(B(c(x))))) b#(c(B(c(b(c(B(x))))))) (19)

1.1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.

1.2 Local Confluence Proof

All critical pairs are joinable which can be seen by computing normal forms of all critical pairs.