Certification Problem

Input (TPDB SRS_Relative/Waldmann_19/random-65)

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

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

and S is the following TRS.

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

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
b(a(a(x1))) → c(c(c(x1))) (7)
a(c(a(x1))) → a(c(c(x1))) (8)
a(c(a(x1))) → a(a(b(x1))) (9)
b(c(b(x1))) → b(b(a(x1))) (10)
b(a(c(x1))) → a(b(b(x1))) (11)
c(a(b(x1))) → c(b(b(x1))) (12)

1.1 Closure Under Flat Contexts

Using the flat contexts

{b(☐), a(☐), c(☐)}

We obtain the transformed TRS
a(c(a(x1))) → a(c(c(x1))) (8)
a(c(a(x1))) → a(a(b(x1))) (9)
b(b(a(a(x1)))) → b(c(c(c(x1)))) (13)
a(b(a(a(x1)))) → a(c(c(c(x1)))) (14)
c(b(a(a(x1)))) → c(c(c(c(x1)))) (15)
b(c(b(x1))) → b(b(a(x1))) (10)
c(a(b(x1))) → c(b(b(x1))) (12)
b(b(a(c(x1)))) → b(a(b(b(x1)))) (16)
a(b(a(c(x1)))) → a(a(b(b(x1)))) (17)
c(b(a(c(x1)))) → c(a(b(b(x1)))) (18)

1.1.1 Semantic Labeling

Root-labeling is applied.

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

1.1.1.1 Rule Removal

Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(ac) = 4 weight(ac) = 25
prec(ca) = 1 weight(ca) = 7
prec(aa) = 2 weight(aa) = 22
prec(cc) = 3 weight(cc) = 11
prec(ab) = 7 weight(ab) = 28
prec(cb) = 8 weight(cb) = 23
prec(ba) = 5 weight(ba) = 2
prec(bc) = 6 weight(bc) = 7
prec(bb) = 0 weight(bb) = 1
all of the following rules can be deleted.
ac(ca(aa(x1))) → ac(cc(ca(x1))) (19)
ac(ca(ac(x1))) → ac(cc(cc(x1))) (20)
ac(ca(ab(x1))) → ac(cc(cb(x1))) (21)
ac(ca(aa(x1))) → aa(ab(ba(x1))) (22)
ac(ca(ac(x1))) → aa(ab(bc(x1))) (23)
ac(ca(ab(x1))) → aa(ab(bb(x1))) (24)
bb(ba(aa(aa(x1)))) → bc(cc(cc(ca(x1)))) (25)
bb(ba(aa(ac(x1)))) → bc(cc(cc(cc(x1)))) (26)
bb(ba(aa(ab(x1)))) → bc(cc(cc(cb(x1)))) (27)
ab(ba(aa(aa(x1)))) → ac(cc(cc(ca(x1)))) (28)
ab(ba(aa(ac(x1)))) → ac(cc(cc(cc(x1)))) (29)
ab(ba(aa(ab(x1)))) → ac(cc(cc(cb(x1)))) (30)
cb(ba(aa(aa(x1)))) → cc(cc(cc(ca(x1)))) (31)
cb(ba(aa(ac(x1)))) → cc(cc(cc(cc(x1)))) (32)
cb(ba(aa(ab(x1)))) → cc(cc(cc(cb(x1)))) (33)
bc(cb(ba(x1))) → bb(ba(aa(x1))) (34)
bc(cb(bc(x1))) → bb(ba(ac(x1))) (35)
bc(cb(bb(x1))) → bb(ba(ab(x1))) (36)
ca(ab(ba(x1))) → cb(bb(ba(x1))) (37)
ca(ab(bc(x1))) → cb(bb(bc(x1))) (38)
ca(ab(bb(x1))) → cb(bb(bb(x1))) (39)
bb(ba(ac(ca(x1)))) → ba(ab(bb(ba(x1)))) (40)
bb(ba(ac(cc(x1)))) → ba(ab(bb(bc(x1)))) (41)
bb(ba(ac(cb(x1)))) → ba(ab(bb(bb(x1)))) (42)
ab(ba(ac(ca(x1)))) → aa(ab(bb(ba(x1)))) (43)
ab(ba(ac(cc(x1)))) → aa(ab(bb(bc(x1)))) (44)
ab(ba(ac(cb(x1)))) → aa(ab(bb(bb(x1)))) (45)
cb(ba(ac(ca(x1)))) → ca(ab(bb(ba(x1)))) (46)
cb(ba(ac(cc(x1)))) → ca(ab(bb(bc(x1)))) (47)
cb(ba(ac(cb(x1)))) → ca(ab(bb(bb(x1)))) (48)

1.1.1.1.1 R is empty

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