Certification Problem

Input (TPDB SRS_Standard/Secret_06_SRS/3-matchbox)

The rewrite relation of the following TRS is considered.

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

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(a(b(c(x1)))) c(b(a(a(x1)))) (5)
a(a(a(b(x1)))) b(a(a(a(x1)))) (6)
c(b(a(x1))) a(b(c(x1))) (7)
b(b(c(c(x1)))) c(c(b(b(x1)))) (8)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
a#(a(b(c(x1)))) c#(b(a(a(x1)))) (9)
a#(a(b(c(x1)))) b#(a(a(x1))) (10)
a#(a(b(c(x1)))) a#(a(x1)) (11)
a#(a(b(c(x1)))) a#(x1) (12)
a#(a(a(b(x1)))) b#(a(a(a(x1)))) (13)
a#(a(a(b(x1)))) a#(a(a(x1))) (14)
a#(a(a(b(x1)))) a#(a(x1)) (15)
a#(a(a(b(x1)))) a#(x1) (16)
c#(b(a(x1))) a#(b(c(x1))) (17)
c#(b(a(x1))) b#(c(x1)) (18)
c#(b(a(x1))) c#(x1) (19)
b#(b(c(c(x1)))) c#(c(b(b(x1)))) (20)
b#(b(c(c(x1)))) c#(b(b(x1))) (21)
b#(b(c(c(x1)))) b#(b(x1)) (22)
b#(b(c(c(x1)))) b#(x1) (23)

1.1.1 Reduction Pair Processor

Using the linear polynomial interpretation over the naturals
[a#(x1)] = 1 + 1 · x1
[a(x1)] = 1 + 1 · x1
[b(x1)] = 1 · x1
[c(x1)] = 1 · x1
[c#(x1)] = 1 · x1
[b#(x1)] = 1 · x1
the pairs
a#(a(b(c(x1)))) a#(x1) (12)
a#(a(a(b(x1)))) a#(a(x1)) (15)
a#(a(a(b(x1)))) a#(x1) (16)
c#(b(a(x1))) b#(c(x1)) (18)
c#(b(a(x1))) c#(x1) (19)
could be deleted.

1.1.1.1 Monotonic Reduction Pair Processor

Using the linear polynomial interpretation over the naturals
[a(x1)] = 1 · x1
[b(x1)] = 1 · x1
[c(x1)] = 2 + 2 · x1
[a#(x1)] = 1 · x1
[c#(x1)] = 2 + 2 · x1
[b#(x1)] = 1 · x1
the pairs
a#(a(b(c(x1)))) b#(a(a(x1))) (10)
a#(a(b(c(x1)))) a#(a(x1)) (11)
b#(b(c(c(x1)))) c#(b(b(x1))) (21)
b#(b(c(c(x1)))) b#(b(x1)) (22)
b#(b(c(c(x1)))) b#(x1) (23)
and no rules could be deleted.

1.1.1.1.1 Reduction Pair Processor

Using the linear polynomial interpretation over the naturals
[a#(x1)] = 1 · x1
[a(x1)] = 1 · x1
[b(x1)] = 1 + 1 · x1
[c(x1)] = 1 · x1
[c#(x1)] = 1 · x1
[b#(x1)] = 1 + 1 · x1
the pair
a#(a(a(b(x1)))) a#(a(a(x1))) (14)
could be deleted.

1.1.1.1.1.1 Reduction Pair Processor

Using the linear polynomial interpretation over the naturals
[a#(x1)] = 2
[b#(x1)] = 1 + x1
[c#(x1)] = 2
[c(x1)] = 1 + x1
[b(x1)] = -2 + 2 · x1
[a(x1)] = 1
the pair
b#(b(c(c(x1)))) c#(c(b(b(x1)))) (20)
could be deleted.

1.1.1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.