Certification Problem

Input (TPDB SRS_Relative/Waldmann_19/random-134)

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

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

and S is the following TRS.

a(c(b(x1))) → c(c(b(x1))) (5)
b(c(b(x1))) → a(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
c(a(a(x1))) → c(c(c(x1))) (7)
b(c(a(x1))) → c(a(a(x1))) (8)
a(c(b(x1))) → b(a(b(x1))) (9)
c(b(a(x1))) → c(a(b(x1))) (10)
b(c(a(x1))) → b(c(c(x1))) (11)
b(c(b(x1))) → c(b(a(x1))) (12)

1.1 Closure Under Flat Contexts

Using the flat contexts

{c(☐), a(☐), b(☐)}

We obtain the transformed TRS
c(a(a(x1))) → c(c(c(x1))) (7)
c(b(a(x1))) → c(a(b(x1))) (10)
c(b(c(a(x1)))) → c(c(a(a(x1)))) (13)
a(b(c(a(x1)))) → a(c(a(a(x1)))) (14)
b(b(c(a(x1)))) → b(c(a(a(x1)))) (15)
c(a(c(b(x1)))) → c(b(a(b(x1)))) (16)
a(a(c(b(x1)))) → a(b(a(b(x1)))) (17)
b(a(c(b(x1)))) → b(b(a(b(x1)))) (18)
b(c(a(x1))) → b(c(c(x1))) (11)
c(b(c(b(x1)))) → c(c(b(a(x1)))) (19)
a(b(c(b(x1)))) → a(c(b(a(x1)))) (20)
b(b(c(b(x1)))) → b(c(b(a(x1)))) (21)

1.1.1 Semantic Labeling

Root-labeling is applied.

We obtain the labeled TRS
ca(aa(ac(x1))) → cc(cc(cc(x1))) (22)
ca(aa(aa(x1))) → cc(cc(ca(x1))) (23)
ca(aa(ab(x1))) → cc(cc(cb(x1))) (24)
cb(ba(ac(x1))) → ca(ab(bc(x1))) (25)
cb(ba(aa(x1))) → ca(ab(ba(x1))) (26)
cb(ba(ab(x1))) → ca(ab(bb(x1))) (27)
cb(bc(ca(ac(x1)))) → cc(ca(aa(ac(x1)))) (28)
cb(bc(ca(aa(x1)))) → cc(ca(aa(aa(x1)))) (29)
cb(bc(ca(ab(x1)))) → cc(ca(aa(ab(x1)))) (30)
ab(bc(ca(ac(x1)))) → ac(ca(aa(ac(x1)))) (31)
ab(bc(ca(aa(x1)))) → ac(ca(aa(aa(x1)))) (32)
ab(bc(ca(ab(x1)))) → ac(ca(aa(ab(x1)))) (33)
bb(bc(ca(ac(x1)))) → bc(ca(aa(ac(x1)))) (34)
bb(bc(ca(aa(x1)))) → bc(ca(aa(aa(x1)))) (35)
bb(bc(ca(ab(x1)))) → bc(ca(aa(ab(x1)))) (36)
ca(ac(cb(bc(x1)))) → cb(ba(ab(bc(x1)))) (37)
ca(ac(cb(ba(x1)))) → cb(ba(ab(ba(x1)))) (38)
ca(ac(cb(bb(x1)))) → cb(ba(ab(bb(x1)))) (39)
aa(ac(cb(bc(x1)))) → ab(ba(ab(bc(x1)))) (40)
aa(ac(cb(ba(x1)))) → ab(ba(ab(ba(x1)))) (41)
aa(ac(cb(bb(x1)))) → ab(ba(ab(bb(x1)))) (42)
ba(ac(cb(bc(x1)))) → bb(ba(ab(bc(x1)))) (43)
ba(ac(cb(ba(x1)))) → bb(ba(ab(ba(x1)))) (44)
ba(ac(cb(bb(x1)))) → bb(ba(ab(bb(x1)))) (45)
bc(ca(ac(x1))) → bc(cc(cc(x1))) (46)
bc(ca(aa(x1))) → bc(cc(ca(x1))) (47)
bc(ca(ab(x1))) → bc(cc(cb(x1))) (48)
cb(bc(cb(bc(x1)))) → cc(cb(ba(ac(x1)))) (49)
cb(bc(cb(ba(x1)))) → cc(cb(ba(aa(x1)))) (50)
cb(bc(cb(bb(x1)))) → cc(cb(ba(ab(x1)))) (51)
ab(bc(cb(bc(x1)))) → ac(cb(ba(ac(x1)))) (52)
ab(bc(cb(ba(x1)))) → ac(cb(ba(aa(x1)))) (53)
ab(bc(cb(bb(x1)))) → ac(cb(ba(ab(x1)))) (54)
bb(bc(cb(bc(x1)))) → bc(cb(ba(ac(x1)))) (55)
bb(bc(cb(ba(x1)))) → bc(cb(ba(aa(x1)))) (56)
bb(bc(cb(bb(x1)))) → bc(cb(ba(ab(x1)))) (57)

1.1.1.1 Rule Removal

Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(ca) = 3 weight(ca) = 6
prec(aa) = 2 weight(aa) = 3
prec(ac) = 8 weight(ac) = 9
prec(cc) = 1 weight(cc) = 1
prec(ab) = 4 weight(ab) = 12
prec(cb) = 0 weight(cb) = 16
prec(ba) = 7 weight(ba) = 2
prec(bc) = 5 weight(bc) = 6
prec(bb) = 6 weight(bb) = 7
all of the following rules can be deleted.
ca(aa(ac(x1))) → cc(cc(cc(x1))) (22)
ca(aa(aa(x1))) → cc(cc(ca(x1))) (23)
ca(aa(ab(x1))) → cc(cc(cb(x1))) (24)
cb(ba(ac(x1))) → ca(ab(bc(x1))) (25)
cb(ba(aa(x1))) → ca(ab(ba(x1))) (26)
cb(ba(ab(x1))) → ca(ab(bb(x1))) (27)
cb(bc(ca(ac(x1)))) → cc(ca(aa(ac(x1)))) (28)
cb(bc(ca(aa(x1)))) → cc(ca(aa(aa(x1)))) (29)
cb(bc(ca(ab(x1)))) → cc(ca(aa(ab(x1)))) (30)
ab(bc(ca(ac(x1)))) → ac(ca(aa(ac(x1)))) (31)
ab(bc(ca(aa(x1)))) → ac(ca(aa(aa(x1)))) (32)
ab(bc(ca(ab(x1)))) → ac(ca(aa(ab(x1)))) (33)
bb(bc(ca(ac(x1)))) → bc(ca(aa(ac(x1)))) (34)
bb(bc(ca(aa(x1)))) → bc(ca(aa(aa(x1)))) (35)
bb(bc(ca(ab(x1)))) → bc(ca(aa(ab(x1)))) (36)
ca(ac(cb(bc(x1)))) → cb(ba(ab(bc(x1)))) (37)
ca(ac(cb(ba(x1)))) → cb(ba(ab(ba(x1)))) (38)
ca(ac(cb(bb(x1)))) → cb(ba(ab(bb(x1)))) (39)
aa(ac(cb(bc(x1)))) → ab(ba(ab(bc(x1)))) (40)
aa(ac(cb(ba(x1)))) → ab(ba(ab(ba(x1)))) (41)
aa(ac(cb(bb(x1)))) → ab(ba(ab(bb(x1)))) (42)
ba(ac(cb(bc(x1)))) → bb(ba(ab(bc(x1)))) (43)
ba(ac(cb(ba(x1)))) → bb(ba(ab(ba(x1)))) (44)
ba(ac(cb(bb(x1)))) → bb(ba(ab(bb(x1)))) (45)
bc(ca(ac(x1))) → bc(cc(cc(x1))) (46)
bc(ca(aa(x1))) → bc(cc(ca(x1))) (47)
bc(ca(ab(x1))) → bc(cc(cb(x1))) (48)
cb(bc(cb(bc(x1)))) → cc(cb(ba(ac(x1)))) (49)
cb(bc(cb(ba(x1)))) → cc(cb(ba(aa(x1)))) (50)
cb(bc(cb(bb(x1)))) → cc(cb(ba(ab(x1)))) (51)
ab(bc(cb(bc(x1)))) → ac(cb(ba(ac(x1)))) (52)
ab(bc(cb(ba(x1)))) → ac(cb(ba(aa(x1)))) (53)
ab(bc(cb(bb(x1)))) → ac(cb(ba(ab(x1)))) (54)
bb(bc(cb(bc(x1)))) → bc(cb(ba(ac(x1)))) (55)
bb(bc(cb(ba(x1)))) → bc(cb(ba(aa(x1)))) (56)
bb(bc(cb(bb(x1)))) → bc(cb(ba(ab(x1)))) (57)

1.1.1.1.1 R is empty

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