Certification Problem

Input (TPDB SRS_Relative/Waldmann_19/random-137)

The rewrite relation of the following TRS is considered.

a(c(a(x1))) → b(a(c(x1))) (1)
a(c(c(x1))) → a(b(b(x1))) (2)
c(a(a(x1))) → a(a(a(x1))) (3)
b(c(c(x1))) → a(c(a(x1))) (4)
b(a(a(x1))) → b(c(b(x1))) (5)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Closure Under Flat Contexts

Using the flat contexts

{a(☐), c(☐), b(☐)}

We obtain the transformed TRS
a(c(c(x1))) → a(b(b(x1))) (2)
b(a(a(x1))) → b(c(b(x1))) (5)
a(a(c(a(x1)))) → a(b(a(c(x1)))) (6)
c(a(c(a(x1)))) → c(b(a(c(x1)))) (7)
b(a(c(a(x1)))) → b(b(a(c(x1)))) (8)
a(c(a(a(x1)))) → a(a(a(a(x1)))) (9)
c(c(a(a(x1)))) → c(a(a(a(x1)))) (10)
b(c(a(a(x1)))) → b(a(a(a(x1)))) (11)
a(b(c(c(x1)))) → a(a(c(a(x1)))) (12)
c(b(c(c(x1)))) → c(a(c(a(x1)))) (13)
b(b(c(c(x1)))) → b(a(c(a(x1)))) (14)

1.1 Semantic Labeling

Root-labeling is applied.

We obtain the labeled TRS
ac(cc(ca(x1))) → ab(bb(ba(x1))) (15)
ac(cc(cc(x1))) → ab(bb(bc(x1))) (16)
ac(cc(cb(x1))) → ab(bb(bb(x1))) (17)
ba(aa(aa(x1))) → bc(cb(ba(x1))) (18)
ba(aa(ac(x1))) → bc(cb(bc(x1))) (19)
ba(aa(ab(x1))) → bc(cb(bb(x1))) (20)
aa(ac(ca(aa(x1)))) → ab(ba(ac(ca(x1)))) (21)
aa(ac(ca(ac(x1)))) → ab(ba(ac(cc(x1)))) (22)
aa(ac(ca(ab(x1)))) → ab(ba(ac(cb(x1)))) (23)
ca(ac(ca(aa(x1)))) → cb(ba(ac(ca(x1)))) (24)
ca(ac(ca(ac(x1)))) → cb(ba(ac(cc(x1)))) (25)
ca(ac(ca(ab(x1)))) → cb(ba(ac(cb(x1)))) (26)
ba(ac(ca(aa(x1)))) → bb(ba(ac(ca(x1)))) (27)
ba(ac(ca(ac(x1)))) → bb(ba(ac(cc(x1)))) (28)
ba(ac(ca(ab(x1)))) → bb(ba(ac(cb(x1)))) (29)
ac(ca(aa(aa(x1)))) → aa(aa(aa(aa(x1)))) (30)
ac(ca(aa(ac(x1)))) → aa(aa(aa(ac(x1)))) (31)
ac(ca(aa(ab(x1)))) → aa(aa(aa(ab(x1)))) (32)
cc(ca(aa(aa(x1)))) → ca(aa(aa(aa(x1)))) (33)
cc(ca(aa(ac(x1)))) → ca(aa(aa(ac(x1)))) (34)
cc(ca(aa(ab(x1)))) → ca(aa(aa(ab(x1)))) (35)
bc(ca(aa(aa(x1)))) → ba(aa(aa(aa(x1)))) (36)
bc(ca(aa(ac(x1)))) → ba(aa(aa(ac(x1)))) (37)
bc(ca(aa(ab(x1)))) → ba(aa(aa(ab(x1)))) (38)
ab(bc(cc(ca(x1)))) → aa(ac(ca(aa(x1)))) (39)
ab(bc(cc(cc(x1)))) → aa(ac(ca(ac(x1)))) (40)
ab(bc(cc(cb(x1)))) → aa(ac(ca(ab(x1)))) (41)
cb(bc(cc(ca(x1)))) → ca(ac(ca(aa(x1)))) (42)
cb(bc(cc(cc(x1)))) → ca(ac(ca(ac(x1)))) (43)
cb(bc(cc(cb(x1)))) → ca(ac(ca(ab(x1)))) (44)
bb(bc(cc(ca(x1)))) → ba(ac(ca(aa(x1)))) (45)
bb(bc(cc(cc(x1)))) → ba(ac(ca(ac(x1)))) (46)
bb(bc(cc(cb(x1)))) → ba(ac(ca(ab(x1)))) (47)

1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[ac(x1)] = 1 · x1
[cc(x1)] = 1 · x1 + 9
[ca(x1)] = 1 · x1 + 10
[ab(x1)] = 1 · x1 + 2
[bb(x1)] = 1 · x1
[ba(x1)] = 1 · x1 + 3
[bc(x1)] = 1 · x1
[cb(x1)] = 1 · x1 + 7
[aa(x1)] = 1 · x1 + 5
all of the following rules can be deleted.
ac(cc(ca(x1))) → ab(bb(ba(x1))) (15)
ac(cc(cc(x1))) → ab(bb(bc(x1))) (16)
ac(cc(cb(x1))) → ab(bb(bb(x1))) (17)
ba(aa(aa(x1))) → bc(cb(ba(x1))) (18)
ba(aa(ac(x1))) → bc(cb(bc(x1))) (19)
ba(aa(ab(x1))) → bc(cb(bb(x1))) (20)
aa(ac(ca(aa(x1)))) → ab(ba(ac(ca(x1)))) (21)
aa(ac(ca(ac(x1)))) → ab(ba(ac(cc(x1)))) (22)
aa(ac(ca(ab(x1)))) → ab(ba(ac(cb(x1)))) (23)
ca(ac(ca(aa(x1)))) → cb(ba(ac(ca(x1)))) (24)
ca(ac(ca(ac(x1)))) → cb(ba(ac(cc(x1)))) (25)
ca(ac(ca(ab(x1)))) → cb(ba(ac(cb(x1)))) (26)
ba(ac(ca(aa(x1)))) → bb(ba(ac(ca(x1)))) (27)
ba(ac(ca(ac(x1)))) → bb(ba(ac(cc(x1)))) (28)
ba(ac(ca(ab(x1)))) → bb(ba(ac(cb(x1)))) (29)
cc(ca(aa(aa(x1)))) → ca(aa(aa(aa(x1)))) (33)
cc(ca(aa(ac(x1)))) → ca(aa(aa(ac(x1)))) (34)
cc(ca(aa(ab(x1)))) → ca(aa(aa(ab(x1)))) (35)
bc(ca(aa(aa(x1)))) → ba(aa(aa(aa(x1)))) (36)
bc(ca(aa(ac(x1)))) → ba(aa(aa(ac(x1)))) (37)
bc(ca(aa(ab(x1)))) → ba(aa(aa(ab(x1)))) (38)
ab(bc(cc(ca(x1)))) → aa(ac(ca(aa(x1)))) (39)
ab(bc(cc(cc(x1)))) → aa(ac(ca(ac(x1)))) (40)
ab(bc(cc(cb(x1)))) → aa(ac(ca(ab(x1)))) (41)
cb(bc(cc(ca(x1)))) → ca(ac(ca(aa(x1)))) (42)
cb(bc(cc(cc(x1)))) → ca(ac(ca(ac(x1)))) (43)
cb(bc(cc(cb(x1)))) → ca(ac(ca(ab(x1)))) (44)
bb(bc(cc(ca(x1)))) → ba(ac(ca(aa(x1)))) (45)
bb(bc(cc(cc(x1)))) → ba(ac(ca(ac(x1)))) (46)
bb(bc(cc(cb(x1)))) → ba(ac(ca(ab(x1)))) (47)

1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[ac(x1)] = 1 · x1
[ca(x1)] = 1 · x1 + 1
[aa(x1)] = 1 · x1
[ab(x1)] = 1 · x1
all of the following rules can be deleted.
ac(ca(aa(aa(x1)))) → aa(aa(aa(aa(x1)))) (30)
ac(ca(aa(ac(x1)))) → aa(aa(aa(ac(x1)))) (31)
ac(ca(aa(ab(x1)))) → aa(aa(aa(ab(x1)))) (32)

1.1.1.1.1 R is empty

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