Certification Problem

Input (TPDB SRS_Relative/Waldmann_19/random-86)

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

c(c(a(x1))) → b(a(c(x1))) (1)
b(b(b(x1))) → a(a(a(x1))) (2)

and S is the following TRS.

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

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS
a(c(c(x1))) → c(a(b(x1))) (7)
b(b(b(x1))) → a(a(a(x1))) (2)
b(a(c(x1))) → a(a(a(x1))) (8)
b(a(c(x1))) → a(c(a(x1))) (9)
c(b(c(x1))) → b(b(c(x1))) (10)
b(a(b(x1))) → c(b(b(x1))) (11)

1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[c(x1)] =
1 1 0
0 0 0
0 0 1
· x1 +
1 0 0
0 0 0
0 0 0
[a(x1)] =
1 0 0
0 0 1
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[b(x1)] =
1 1 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
1 0 0
all of the following rules can be deleted.
b(a(c(x1))) → a(a(a(x1))) (8)
c(b(c(x1))) → b(b(c(x1))) (10)

1.1.1 Rule Removal

Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(b) = 3 weight(b) = 2
prec(c) = 0 weight(c) = 2
prec(a) = 1 weight(a) = 2
all of the following rules can be deleted.
a(c(c(x1))) → c(a(b(x1))) (7)
b(b(b(x1))) → a(a(a(x1))) (2)
b(a(c(x1))) → a(c(a(x1))) (9)
b(a(b(x1))) → c(b(b(x1))) (11)

1.1.1.1 R is empty

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