Certification Problem

Input (TPDB SRS_Standard/Trafo_06/dup06)

The rewrite relation of the following TRS is considered.

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

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 Rule Removal

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

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
b#(b(c(c(x1)))) b#(x1) (4)
b#(b(c(c(x1)))) b#(b(x1)) (5)
b#(b(c(c(x1)))) b#(b(b(x1))) (6)
b#(b(c(c(x1)))) b#(b(b(b(x1)))) (7)
b#(b(a(a(x1)))) b#(x1) (8)
b#(b(a(a(x1)))) b#(b(x1)) (9)

1.1.1 Subterm Criterion Processor

We use the projection to multisets
π(b#) = { 1, 1 }
π(c) = { 1 }
π(a) = { 1, 1 }
π(b) = { 1 }
to remove the pairs:
b#(b(a(a(x1)))) b#(x1) (8)
b#(b(a(a(x1)))) b#(b(x1)) (9)

1.1.1.1 Reduction Pair Processor with Usable Rules

Using the
prec(b#) = 0 stat(b#) = lex
prec(c) = 0 stat(c) = lex
prec(a) = 0 stat(a) = lex
prec(b) = 0 stat(b) = lex

π(b#) = 1
π(c) = [1]
π(a) = []
π(b) = 1

together with the usable rules
b(b(c(c(x1)))) c(c(b(b(b(b(x1)))))) (2)
b(b(a(a(x1)))) a(a(c(c(b(b(x1)))))) (3)
(w.r.t. the implicit argument filter of the reduction pair), the pairs
b#(b(c(c(x1)))) b#(x1) (4)
b#(b(c(c(x1)))) b#(b(x1)) (5)
b#(b(c(c(x1)))) b#(b(b(x1))) (6)
b#(b(c(c(x1)))) b#(b(b(b(x1)))) (7)
could be deleted.

1.1.1.1.1 P is empty

There are no pairs anymore.