Certification Problem

Input (TPDB SRS_Relative/Waldmann_19/random-151)

The relative rewrite relation R/S is considered where R is the following TRS

c(b(a(x1))) → c(b(b(x1))) (1)
b(a(c(x1))) → b(b(c(x1))) (2)
c(c(c(x1))) → b(c(c(x1))) (3)
a(c(a(x1))) → b(c(b(x1))) (4)
b(c(b(x1))) → c(b(c(x1))) (5)
a(c(b(x1))) → b(a(c(x1))) (6)

and S is the following TRS.

b(b(a(x1))) → c(b(b(x1))) (7)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS
a(b(c(x1))) → b(b(c(x1))) (8)
c(a(b(x1))) → c(b(b(x1))) (9)
c(c(c(x1))) → c(c(b(x1))) (10)
a(c(a(x1))) → b(c(b(x1))) (4)
b(c(b(x1))) → c(b(c(x1))) (5)
b(c(a(x1))) → c(a(b(x1))) (11)
a(b(b(x1))) → b(b(c(x1))) (12)

1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[a(x1)] = 1 + 1 · x1
[b(x1)] = 1 · x1
[c(x1)] = 1 · x1
all of the following rules can be deleted.
a(b(c(x1))) → b(b(c(x1))) (8)
c(a(b(x1))) → c(b(b(x1))) (9)
a(c(a(x1))) → b(c(b(x1))) (4)
a(b(b(x1))) → b(b(c(x1))) (12)

1.1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
c#(c(c(x1))) → c#(c(b(x1))) (13)
c#(c(c(x1))) → c#(b(x1)) (14)
c#(c(c(x1))) → b#(x1) (15)
b#(c(b(x1))) → c#(b(c(x1))) (16)
b#(c(b(x1))) → b#(c(x1)) (17)
b#(c(b(x1))) → c#(x1) (18)
b#(c(a(x1))) → c#(a(b(x1))) (19)
b#(c(a(x1))) → b#(x1) (20)

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.