Certification Problem

Input (TPDB SRS_Standard/Trafo_06/dup08)

The rewrite relation of the following TRS is considered.

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

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Closure Under Flat Contexts

Using the flat contexts

{a(☐), b(☐), c(☐)}

We obtain the transformed TRS
a(a(a(a(a(b(b(x1))))))) → a(b(b(a(a(b(b(x1))))))) (4)
b(a(a(a(a(b(b(x1))))))) → b(b(b(a(a(b(b(x1))))))) (5)
c(a(a(a(a(b(b(x1))))))) → c(b(b(a(a(b(b(x1))))))) (6)
a(b(b(a(a(x1))))) → a(a(a(b(b(b(b(x1))))))) (7)
b(b(b(a(a(x1))))) → b(a(a(b(b(b(b(x1))))))) (8)
c(b(b(a(a(x1))))) → c(a(a(b(b(b(b(x1))))))) (9)
a(b(b(c(c(a(a(x1))))))) → a(c(c(c(c(a(a(a(a(b(b(x1))))))))))) (10)
b(b(b(c(c(a(a(x1))))))) → b(c(c(c(c(a(a(a(a(b(b(x1))))))))))) (11)
c(b(b(c(c(a(a(x1))))))) → c(c(c(c(c(a(a(a(a(b(b(x1))))))))))) (12)

1.1 Semantic Labeling

Root-labeling is applied.

We obtain the labeled TRS
aa(aa(aa(aa(ab(bb(ba(x1))))))) → ab(bb(ba(aa(ab(bb(ba(x1))))))) (13)
aa(aa(aa(aa(ab(bb(bb(x1))))))) → ab(bb(ba(aa(ab(bb(bb(x1))))))) (14)
aa(aa(aa(aa(ab(bb(bc(x1))))))) → ab(bb(ba(aa(ab(bb(bc(x1))))))) (15)
ba(aa(aa(aa(ab(bb(ba(x1))))))) → bb(bb(ba(aa(ab(bb(ba(x1))))))) (16)
ba(aa(aa(aa(ab(bb(bb(x1))))))) → bb(bb(ba(aa(ab(bb(bb(x1))))))) (17)
ba(aa(aa(aa(ab(bb(bc(x1))))))) → bb(bb(ba(aa(ab(bb(bc(x1))))))) (18)
ca(aa(aa(aa(ab(bb(ba(x1))))))) → cb(bb(ba(aa(ab(bb(ba(x1))))))) (19)
ca(aa(aa(aa(ab(bb(bb(x1))))))) → cb(bb(ba(aa(ab(bb(bb(x1))))))) (20)
ca(aa(aa(aa(ab(bb(bc(x1))))))) → cb(bb(ba(aa(ab(bb(bc(x1))))))) (21)
ab(bb(ba(aa(aa(x1))))) → aa(aa(ab(bb(bb(bb(ba(x1))))))) (22)
ab(bb(ba(aa(ab(x1))))) → aa(aa(ab(bb(bb(bb(bb(x1))))))) (23)
ab(bb(ba(aa(ac(x1))))) → aa(aa(ab(bb(bb(bb(bc(x1))))))) (24)
bb(bb(ba(aa(aa(x1))))) → ba(aa(ab(bb(bb(bb(ba(x1))))))) (25)
bb(bb(ba(aa(ab(x1))))) → ba(aa(ab(bb(bb(bb(bb(x1))))))) (26)
bb(bb(ba(aa(ac(x1))))) → ba(aa(ab(bb(bb(bb(bc(x1))))))) (27)
cb(bb(ba(aa(aa(x1))))) → ca(aa(ab(bb(bb(bb(ba(x1))))))) (28)
cb(bb(ba(aa(ab(x1))))) → ca(aa(ab(bb(bb(bb(bb(x1))))))) (29)
cb(bb(ba(aa(ac(x1))))) → ca(aa(ab(bb(bb(bb(bc(x1))))))) (30)
ab(bb(bc(cc(ca(aa(aa(x1))))))) → ac(cc(cc(cc(ca(aa(aa(aa(ab(bb(ba(x1))))))))))) (31)
ab(bb(bc(cc(ca(aa(ab(x1))))))) → ac(cc(cc(cc(ca(aa(aa(aa(ab(bb(bb(x1))))))))))) (32)
ab(bb(bc(cc(ca(aa(ac(x1))))))) → ac(cc(cc(cc(ca(aa(aa(aa(ab(bb(bc(x1))))))))))) (33)
bb(bb(bc(cc(ca(aa(aa(x1))))))) → bc(cc(cc(cc(ca(aa(aa(aa(ab(bb(ba(x1))))))))))) (34)
bb(bb(bc(cc(ca(aa(ab(x1))))))) → bc(cc(cc(cc(ca(aa(aa(aa(ab(bb(bb(x1))))))))))) (35)
bb(bb(bc(cc(ca(aa(ac(x1))))))) → bc(cc(cc(cc(ca(aa(aa(aa(ab(bb(bc(x1))))))))))) (36)
cb(bb(bc(cc(ca(aa(aa(x1))))))) → cc(cc(cc(cc(ca(aa(aa(aa(ab(bb(ba(x1))))))))))) (37)
cb(bb(bc(cc(ca(aa(ab(x1))))))) → cc(cc(cc(cc(ca(aa(aa(aa(ab(bb(bb(x1))))))))))) (38)
cb(bb(bc(cc(ca(aa(ac(x1))))))) → cc(cc(cc(cc(ca(aa(aa(aa(ab(bb(bc(x1))))))))))) (39)

1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[aa(x1)] = 1 · x1
[ab(x1)] = 1 · x1
[bb(x1)] = 1 · x1
[ba(x1)] = 1 · x1
[bc(x1)] = 1 · x1 + 1
[ca(x1)] = 1 · x1
[cb(x1)] = 1 · x1
[ac(x1)] = 1 · x1 + 1
[cc(x1)] = 1 · x1
all of the following rules can be deleted.
cb(bb(bc(cc(ca(aa(aa(x1))))))) → cc(cc(cc(cc(ca(aa(aa(aa(ab(bb(ba(x1))))))))))) (37)
cb(bb(bc(cc(ca(aa(ab(x1))))))) → cc(cc(cc(cc(ca(aa(aa(aa(ab(bb(bb(x1))))))))))) (38)
cb(bb(bc(cc(ca(aa(ac(x1))))))) → cc(cc(cc(cc(ca(aa(aa(aa(ab(bb(bc(x1))))))))))) (39)

1.1.1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

There are 127 ruless (increase limit for explicit display).

1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.