Certification Problem

Input (TPDB SRS_Relative/Waldmann_19/random-149)

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

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

and S is the following TRS.

a(b(b(x1))) → a(b(b(x1))) (4)
a(b(b(x1))) → a(b(a(x1))) (5)
a(c(a(x1))) → a(c(c(x1))) (6)
b(c(c(x1))) → c(b(c(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(c(a(x1))) → c(a(b(x1))) (8)
b(b(a(x1))) → c(a(a(x1))) (9)
c(a(b(x1))) → a(c(c(x1))) (10)
b(b(a(x1))) → a(b(a(x1))) (11)
a(c(a(x1))) → c(c(a(x1))) (12)
c(c(b(x1))) → c(b(c(x1))) (13)

1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[a(x1)] = 1 + 1 · x1
[c(x1)] = 1 · x1
[b(x1)] = 1 + 1 · x1
all of the following rules can be deleted.
b(b(a(x1))) → c(a(a(x1))) (9)
c(a(b(x1))) → a(c(c(x1))) (10)
a(c(a(x1))) → c(c(a(x1))) (12)

1.1.1 Rule Removal

Using the matrix interpretations of dimension 2 with strict dimension 1 over the integers
[a(x1)] =
0
0
+
1 1
0 2
· x1
[c(x1)] =
0
1
+
1 0
0 0
· x1
[b(x1)] =
0
0
+
1 0
0 0
· x1
all of the following rules can be deleted.
a(c(a(x1))) → c(a(b(x1))) (8)

1.1.1.1 R is empty

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