Certification Problem

Input (TPDB SRS_Standard/Trafo_06/hom01)

The rewrite relation of the following TRS is considered.

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

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

1.1 Rule Removal

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

1.1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[d(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[c(x1)] =
1 1 0
0 1 0
0 0 1
· x1 +
0 0 0
0 0 0
0 0 0
[a(x1)] =
1 0 0
0 1 1
0 1 0
· x1 +
0 0 0
0 0 0
1 0 0
[b(x1)] =
1 1 0
1 0 0
0 0 0
· x1 +
0 0 0
0 0 0
1 0 0
all of the following rules can be deleted.
c(a(a(x1))) → a(a(c(c(x1)))) (5)

1.1.1.1 Rule Removal

Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(c) = 0 weight(c) = 4
prec(b) = 2 weight(b) = 1
prec(d) = 1 weight(d) = 1
all of the following rules can be deleted.
c(c(c(x1))) → d(b(c(d(b(x1))))) (6)

1.1.1.1.1 R is empty

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