Certification Problem

Input (TPDB SRS_Relative/Waldmann_19/random-70)

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

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

and S is the following TRS.

c(a(a(x1))) → c(a(a(x1))) (5)
c(c(a(x1))) → c(b(b(x1))) (6)
a(a(b(x1))) → b(a(b(x1))) (7)

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(a(b(x1))) → b(b(c(x1))) (8)
a(a(a(x1))) → c(c(a(x1))) (9)
b(c(a(x1))) → b(a(a(x1))) (10)
c(a(c(x1))) → c(c(c(x1))) (4)
a(a(c(x1))) → a(a(c(x1))) (11)
a(c(c(x1))) → b(b(c(x1))) (12)
b(a(a(x1))) → b(a(b(x1))) (13)

1.1 Rule Removal

Using the linear polynomial interpretation over the arctic semiring over the integers
[b(x1)] = 0 · x1 + -∞
[a(x1)] = 1 · x1 + -∞
[c(x1)] = 1 · x1 + -∞
all of the following rules can be deleted.
a(a(b(x1))) → b(b(c(x1))) (8)
a(c(c(x1))) → b(b(c(x1))) (12)
b(a(a(x1))) → b(a(b(x1))) (13)

1.1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[b(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[a(x1)] =
1 0 1
1 1 0
0 1 0
· x1 +
0 0 0
1 0 0
0 0 0
[c(x1)] =
1 0 1
0 1 1
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
a(a(a(x1))) → c(c(a(x1))) (9)

1.1.1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[b(x1)] =
1 0 0
1 0 1
0 0 1
· x1 +
1 0 0
0 0 0
1 0 0
[a(x1)] =
1 0 0
0 1 1
0 0 0
· x1 +
0 0 0
1 0 0
0 0 0
[c(x1)] =
1 1 0
0 1 0
0 1 0
· x1 +
0 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
b(c(a(x1))) → b(a(a(x1))) (10)
c(a(c(x1))) → c(c(c(x1))) (4)

1.1.1.1.1 R is empty

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