Certification Problem

Input (TPDB SRS_Relative/Waldmann_19/random-162)

The relative rewrite relation R/S is considered where R is the following TRS

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

and S is the following TRS.

b(c(a(x1))) → c(b(a(x1))) (4)
a(b(a(x1))) → b(a(a(x1))) (5)
a(c(c(x1))) → a(b(c(x1))) (6)

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(a(a(x1))) → a(c(a(x1))) (2)
c(c(b(x1))) → c(a(b(x1))) (3)
a(a(a(a(x1)))) → a(c(b(b(x1)))) (7)
c(a(a(a(x1)))) → c(c(b(b(x1)))) (8)
b(a(a(a(x1)))) → b(c(b(b(x1)))) (9)
a(c(c(x1))) → a(b(c(x1))) (6)
a(b(c(a(x1)))) → a(c(b(a(x1)))) (10)
c(b(c(a(x1)))) → c(c(b(a(x1)))) (11)
b(b(c(a(x1)))) → b(c(b(a(x1)))) (12)
a(a(b(a(x1)))) → a(b(a(a(x1)))) (13)
c(a(b(a(x1)))) → c(b(a(a(x1)))) (14)
b(a(b(a(x1)))) → b(b(a(a(x1)))) (15)

1.1 Semantic Labeling

Root-labeling is applied.

We obtain the labeled TRS
aa(aa(aa(x1))) → ac(ca(aa(x1))) (16)
aa(aa(ac(x1))) → ac(ca(ac(x1))) (17)
aa(aa(ab(x1))) → ac(ca(ab(x1))) (18)
cc(cb(ba(x1))) → ca(ab(ba(x1))) (19)
cc(cb(bc(x1))) → ca(ab(bc(x1))) (20)
cc(cb(bb(x1))) → ca(ab(bb(x1))) (21)
aa(aa(aa(aa(x1)))) → ac(cb(bb(ba(x1)))) (22)
aa(aa(aa(ac(x1)))) → ac(cb(bb(bc(x1)))) (23)
aa(aa(aa(ab(x1)))) → ac(cb(bb(bb(x1)))) (24)
ca(aa(aa(aa(x1)))) → cc(cb(bb(ba(x1)))) (25)
ca(aa(aa(ac(x1)))) → cc(cb(bb(bc(x1)))) (26)
ca(aa(aa(ab(x1)))) → cc(cb(bb(bb(x1)))) (27)
ba(aa(aa(aa(x1)))) → bc(cb(bb(ba(x1)))) (28)
ba(aa(aa(ac(x1)))) → bc(cb(bb(bc(x1)))) (29)
ba(aa(aa(ab(x1)))) → bc(cb(bb(bb(x1)))) (30)
ac(cc(ca(x1))) → ab(bc(ca(x1))) (31)
ac(cc(cc(x1))) → ab(bc(cc(x1))) (32)
ac(cc(cb(x1))) → ab(bc(cb(x1))) (33)
ab(bc(ca(aa(x1)))) → ac(cb(ba(aa(x1)))) (34)
ab(bc(ca(ac(x1)))) → ac(cb(ba(ac(x1)))) (35)
ab(bc(ca(ab(x1)))) → ac(cb(ba(ab(x1)))) (36)
cb(bc(ca(aa(x1)))) → cc(cb(ba(aa(x1)))) (37)
cb(bc(ca(ac(x1)))) → cc(cb(ba(ac(x1)))) (38)
cb(bc(ca(ab(x1)))) → cc(cb(ba(ab(x1)))) (39)
bb(bc(ca(aa(x1)))) → bc(cb(ba(aa(x1)))) (40)
bb(bc(ca(ac(x1)))) → bc(cb(ba(ac(x1)))) (41)
bb(bc(ca(ab(x1)))) → bc(cb(ba(ab(x1)))) (42)
aa(ab(ba(aa(x1)))) → ab(ba(aa(aa(x1)))) (43)
aa(ab(ba(ac(x1)))) → ab(ba(aa(ac(x1)))) (44)
aa(ab(ba(ab(x1)))) → ab(ba(aa(ab(x1)))) (45)
ca(ab(ba(aa(x1)))) → cb(ba(aa(aa(x1)))) (46)
ca(ab(ba(ac(x1)))) → cb(ba(aa(ac(x1)))) (47)
ca(ab(ba(ab(x1)))) → cb(ba(aa(ab(x1)))) (48)
ba(ab(ba(aa(x1)))) → bb(ba(aa(aa(x1)))) (49)
ba(ab(ba(ac(x1)))) → bb(ba(aa(ac(x1)))) (50)
ba(ab(ba(ab(x1)))) → bb(ba(aa(ab(x1)))) (51)

1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[aa(x1)] = 1 + 1 · x1
[ac(x1)] = 1 + 1 · x1
[ca(x1)] = 1 · x1
[ab(x1)] = 1 + 1 · x1
[cc(x1)] = 1 + 1 · x1
[cb(x1)] = 1 · x1
[ba(x1)] = 1 · x1
[bc(x1)] = 1 + 1 · x1
[bb(x1)] = 1 · x1
all of the following rules can be deleted.
aa(aa(aa(x1))) → ac(ca(aa(x1))) (16)
aa(aa(ac(x1))) → ac(ca(ac(x1))) (17)
aa(aa(ab(x1))) → ac(ca(ab(x1))) (18)
aa(aa(aa(aa(x1)))) → ac(cb(bb(ba(x1)))) (22)
aa(aa(aa(ac(x1)))) → ac(cb(bb(bc(x1)))) (23)
aa(aa(aa(ab(x1)))) → ac(cb(bb(bb(x1)))) (24)
ca(aa(aa(aa(x1)))) → cc(cb(bb(ba(x1)))) (25)
ca(aa(aa(ac(x1)))) → cc(cb(bb(bc(x1)))) (26)
ca(aa(aa(ab(x1)))) → cc(cb(bb(bb(x1)))) (27)
ba(aa(aa(aa(x1)))) → bc(cb(bb(ba(x1)))) (28)
ba(aa(aa(ac(x1)))) → bc(cb(bb(bc(x1)))) (29)
ba(aa(aa(ab(x1)))) → bc(cb(bb(bb(x1)))) (30)
ab(bc(ca(aa(x1)))) → ac(cb(ba(aa(x1)))) (34)
ab(bc(ca(ac(x1)))) → ac(cb(ba(ac(x1)))) (35)
ab(bc(ca(ab(x1)))) → ac(cb(ba(ab(x1)))) (36)

1.1.1.1 Rule Removal

Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(cc) = 0 weight(cc) = 7
prec(cb) = 1 weight(cb) = 8
prec(ba) = 7 weight(ba) = 5
prec(ca) = 2 weight(ca) = 9
prec(ab) = 3 weight(ab) = 5
prec(bc) = 4 weight(bc) = 3
prec(bb) = 5 weight(bb) = 4
prec(ac) = 6 weight(ac) = 1
prec(aa) = 8 weight(aa) = 6
all of the following rules can be deleted.
cc(cb(ba(x1))) → ca(ab(ba(x1))) (19)
cc(cb(bc(x1))) → ca(ab(bc(x1))) (20)
cc(cb(bb(x1))) → ca(ab(bb(x1))) (21)
ac(cc(ca(x1))) → ab(bc(ca(x1))) (31)
ac(cc(cc(x1))) → ab(bc(cc(x1))) (32)
ac(cc(cb(x1))) → ab(bc(cb(x1))) (33)
cb(bc(ca(aa(x1)))) → cc(cb(ba(aa(x1)))) (37)
cb(bc(ca(ac(x1)))) → cc(cb(ba(ac(x1)))) (38)
cb(bc(ca(ab(x1)))) → cc(cb(ba(ab(x1)))) (39)
bb(bc(ca(aa(x1)))) → bc(cb(ba(aa(x1)))) (40)
bb(bc(ca(ac(x1)))) → bc(cb(ba(ac(x1)))) (41)
bb(bc(ca(ab(x1)))) → bc(cb(ba(ab(x1)))) (42)
aa(ab(ba(aa(x1)))) → ab(ba(aa(aa(x1)))) (43)
aa(ab(ba(ac(x1)))) → ab(ba(aa(ac(x1)))) (44)
aa(ab(ba(ab(x1)))) → ab(ba(aa(ab(x1)))) (45)
ca(ab(ba(aa(x1)))) → cb(ba(aa(aa(x1)))) (46)
ca(ab(ba(ac(x1)))) → cb(ba(aa(ac(x1)))) (47)
ca(ab(ba(ab(x1)))) → cb(ba(aa(ab(x1)))) (48)
ba(ab(ba(aa(x1)))) → bb(ba(aa(aa(x1)))) (49)
ba(ab(ba(ac(x1)))) → bb(ba(aa(ac(x1)))) (50)
ba(ab(ba(ab(x1)))) → bb(ba(aa(ab(x1)))) (51)

1.1.1.1.1 R is empty

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