Certification Problem

Input (TPDB SRS_Standard/Trafo_06/dup14)

The rewrite relation of the following TRS is considered.

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

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(b(b(b(a(a(x1)))))))) → b(b(a(a(c(c(a(a(x1)))))))) (4)
c(c(a(a(x1)))) → a(a(c(c(c(c(x1)))))) (5)
c(c(c(c(c(c(x1)))))) → b(b(c(c(b(b(x1)))))) (3)

1.1 Closure Under Flat Contexts

Using the flat contexts

{a(☐), b(☐), c(☐)}

We obtain the transformed TRS
a(a(a(b(b(b(b(a(a(x1))))))))) → a(b(b(a(a(c(c(a(a(x1))))))))) (6)
b(a(a(b(b(b(b(a(a(x1))))))))) → b(b(b(a(a(c(c(a(a(x1))))))))) (7)
c(a(a(b(b(b(b(a(a(x1))))))))) → c(b(b(a(a(c(c(a(a(x1))))))))) (8)
a(c(c(a(a(x1))))) → a(a(a(c(c(c(c(x1))))))) (9)
b(c(c(a(a(x1))))) → b(a(a(c(c(c(c(x1))))))) (10)
c(c(c(a(a(x1))))) → c(a(a(c(c(c(c(x1))))))) (11)
a(c(c(c(c(c(c(x1))))))) → a(b(b(c(c(b(b(x1))))))) (12)
b(c(c(c(c(c(c(x1))))))) → b(b(b(c(c(b(b(x1))))))) (13)
c(c(c(c(c(c(c(x1))))))) → c(b(b(c(c(b(b(x1))))))) (14)

1.1.1 Semantic Labeling

Root-labeling is applied.

We obtain the labeled TRS
aa(aa(ab(bb(bb(bb(ba(aa(aa(x1))))))))) → ab(bb(ba(aa(ac(cc(ca(aa(aa(x1))))))))) (15)
aa(aa(ab(bb(bb(bb(ba(aa(ab(x1))))))))) → ab(bb(ba(aa(ac(cc(ca(aa(ab(x1))))))))) (16)
aa(aa(ab(bb(bb(bb(ba(aa(ac(x1))))))))) → ab(bb(ba(aa(ac(cc(ca(aa(ac(x1))))))))) (17)
ba(aa(ab(bb(bb(bb(ba(aa(aa(x1))))))))) → bb(bb(ba(aa(ac(cc(ca(aa(aa(x1))))))))) (18)
ba(aa(ab(bb(bb(bb(ba(aa(ab(x1))))))))) → bb(bb(ba(aa(ac(cc(ca(aa(ab(x1))))))))) (19)
ba(aa(ab(bb(bb(bb(ba(aa(ac(x1))))))))) → bb(bb(ba(aa(ac(cc(ca(aa(ac(x1))))))))) (20)
ca(aa(ab(bb(bb(bb(ba(aa(aa(x1))))))))) → cb(bb(ba(aa(ac(cc(ca(aa(aa(x1))))))))) (21)
ca(aa(ab(bb(bb(bb(ba(aa(ab(x1))))))))) → cb(bb(ba(aa(ac(cc(ca(aa(ab(x1))))))))) (22)
ca(aa(ab(bb(bb(bb(ba(aa(ac(x1))))))))) → cb(bb(ba(aa(ac(cc(ca(aa(ac(x1))))))))) (23)
ac(cc(ca(aa(aa(x1))))) → aa(aa(ac(cc(cc(cc(ca(x1))))))) (24)
ac(cc(ca(aa(ab(x1))))) → aa(aa(ac(cc(cc(cc(cb(x1))))))) (25)
ac(cc(ca(aa(ac(x1))))) → aa(aa(ac(cc(cc(cc(cc(x1))))))) (26)
bc(cc(ca(aa(aa(x1))))) → ba(aa(ac(cc(cc(cc(ca(x1))))))) (27)
bc(cc(ca(aa(ab(x1))))) → ba(aa(ac(cc(cc(cc(cb(x1))))))) (28)
bc(cc(ca(aa(ac(x1))))) → ba(aa(ac(cc(cc(cc(cc(x1))))))) (29)
cc(cc(ca(aa(aa(x1))))) → ca(aa(ac(cc(cc(cc(ca(x1))))))) (30)
cc(cc(ca(aa(ab(x1))))) → ca(aa(ac(cc(cc(cc(cb(x1))))))) (31)
cc(cc(ca(aa(ac(x1))))) → ca(aa(ac(cc(cc(cc(cc(x1))))))) (32)
ac(cc(cc(cc(cc(cc(ca(x1))))))) → ab(bb(bc(cc(cb(bb(ba(x1))))))) (33)
ac(cc(cc(cc(cc(cc(cb(x1))))))) → ab(bb(bc(cc(cb(bb(bb(x1))))))) (34)
ac(cc(cc(cc(cc(cc(cc(x1))))))) → ab(bb(bc(cc(cb(bb(bc(x1))))))) (35)
bc(cc(cc(cc(cc(cc(ca(x1))))))) → bb(bb(bc(cc(cb(bb(ba(x1))))))) (36)
bc(cc(cc(cc(cc(cc(cb(x1))))))) → bb(bb(bc(cc(cb(bb(bb(x1))))))) (37)
bc(cc(cc(cc(cc(cc(cc(x1))))))) → bb(bb(bc(cc(cb(bb(bc(x1))))))) (38)
cc(cc(cc(cc(cc(cc(ca(x1))))))) → cb(bb(bc(cc(cb(bb(ba(x1))))))) (39)
cc(cc(cc(cc(cc(cc(cb(x1))))))) → cb(bb(bc(cc(cb(bb(bb(x1))))))) (40)
cc(cc(cc(cc(cc(cc(cc(x1))))))) → cb(bb(bc(cc(cb(bb(bc(x1))))))) (41)

1.1.1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

There are 129 ruless (increase limit for explicit display).

1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.