Certification Problem

Input (TPDB SRS_Standard/Waldmann_06_SRS/sym-6)

The rewrite relation of the following TRS is considered.

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

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by matchbox @ termCOMP 2023)

1 Closure Under Flat Contexts

Using the flat contexts

{c(), b()}

We obtain the transformed TRS
c(c(c(c(c(x1))))) c(b(b(b(b(x1))))) (4)
c(b(b(x1))) c(x1) (5)
c(b(b(x1))) c(c(b(c(x1)))) (6)
b(c(c(c(c(x1))))) b(b(b(b(b(x1))))) (7)
b(b(b(x1))) b(x1) (8)
b(b(b(x1))) b(c(b(c(x1)))) (9)

1.1 Semantic Labeling

The following interpretations form a model of the rules.

As carrier we take the set {0,1}. Symbols are labeled by the interpretation of their arguments using the interpretations (modulo 2):

[c(x1)] = 2x1 + 0
[b(x1)] = 2x1 + 1

We obtain the labeled TRS
c0(c0(c0(c0(c0(x1))))) c1(b1(b1(b1(b0(x1))))) (10)
c0(c0(c0(c0(c1(x1))))) c1(b1(b1(b1(b1(x1))))) (11)
b0(c0(c0(c0(c0(x1))))) b1(b1(b1(b1(b0(x1))))) (12)
b0(c0(c0(c0(c1(x1))))) b1(b1(b1(b1(b1(x1))))) (13)
c1(b1(b0(x1))) c0(x1) (14)
c1(b1(b1(x1))) c1(x1) (15)
b1(b1(b0(x1))) b0(x1) (16)
b1(b1(b1(x1))) b1(x1) (17)
c1(b1(b0(x1))) c0(c1(b0(c0(x1)))) (18)
c1(b1(b1(x1))) c0(c1(b0(c1(x1)))) (19)
b1(b1(b0(x1))) b0(c1(b0(c0(x1)))) (20)
b1(b1(b1(x1))) b0(c1(b0(c1(x1)))) (21)

1.1.1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS
c0(c0(c0(c0(c0(x1))))) b0(b1(b1(b1(c1(x1))))) (22)
c1(c0(c0(c0(c0(x1))))) b1(b1(b1(b1(c1(x1))))) (23)
c0(c0(c0(c0(b0(x1))))) b0(b1(b1(b1(b1(x1))))) (24)
c1(c0(c0(c0(b0(x1))))) b1(b1(b1(b1(b1(x1))))) (25)
b0(b1(c1(x1))) c0(x1) (26)
b1(b1(c1(x1))) c1(x1) (27)
b0(b1(b1(x1))) b0(x1) (28)
b1(b1(b1(x1))) b1(x1) (17)
b0(b1(c1(x1))) c0(b0(c1(c0(x1)))) (29)
b1(b1(c1(x1))) c1(b0(c1(c0(x1)))) (30)
b0(b1(b1(x1))) c0(b0(c1(b0(x1)))) (31)
b1(b1(b1(x1))) c1(b0(c1(b0(x1)))) (32)

1.1.1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
c0#(c0(c0(c0(c0(x1))))) c1#(x1) (33)
c0#(c0(c0(c0(c0(x1))))) b0#(b1(b1(b1(c1(x1))))) (34)
c0#(c0(c0(c0(c0(x1))))) b1#(c1(x1)) (35)
c0#(c0(c0(c0(c0(x1))))) b1#(b1(c1(x1))) (36)
c0#(c0(c0(c0(c0(x1))))) b1#(b1(b1(c1(x1)))) (37)
c0#(c0(c0(c0(b0(x1))))) b0#(b1(b1(b1(b1(x1))))) (38)
c0#(c0(c0(c0(b0(x1))))) b1#(x1) (39)
c0#(c0(c0(c0(b0(x1))))) b1#(b1(x1)) (40)
c0#(c0(c0(c0(b0(x1))))) b1#(b1(b1(x1))) (41)
c0#(c0(c0(c0(b0(x1))))) b1#(b1(b1(b1(x1)))) (42)
c1#(c0(c0(c0(c0(x1))))) c1#(x1) (43)
c1#(c0(c0(c0(c0(x1))))) b1#(c1(x1)) (44)
c1#(c0(c0(c0(c0(x1))))) b1#(b1(c1(x1))) (45)
c1#(c0(c0(c0(c0(x1))))) b1#(b1(b1(c1(x1)))) (46)
c1#(c0(c0(c0(c0(x1))))) b1#(b1(b1(b1(c1(x1))))) (47)
c1#(c0(c0(c0(b0(x1))))) b1#(x1) (48)
c1#(c0(c0(c0(b0(x1))))) b1#(b1(x1)) (49)
c1#(c0(c0(c0(b0(x1))))) b1#(b1(b1(x1))) (50)
c1#(c0(c0(c0(b0(x1))))) b1#(b1(b1(b1(x1)))) (51)
c1#(c0(c0(c0(b0(x1))))) b1#(b1(b1(b1(b1(x1))))) (52)
b0#(b1(c1(x1))) c0#(x1) (53)
b0#(b1(c1(x1))) c0#(b0(c1(c0(x1)))) (54)
b0#(b1(c1(x1))) c1#(c0(x1)) (55)
b0#(b1(c1(x1))) b0#(c1(c0(x1))) (56)
b0#(b1(b1(x1))) c0#(b0(c1(b0(x1)))) (57)
b0#(b1(b1(x1))) c1#(b0(x1)) (58)
b0#(b1(b1(x1))) b0#(x1) (59)
b0#(b1(b1(x1))) b0#(c1(b0(x1))) (60)
b1#(b1(c1(x1))) c0#(x1) (61)
b1#(b1(c1(x1))) c1#(c0(x1)) (62)
b1#(b1(c1(x1))) c1#(b0(c1(c0(x1)))) (63)
b1#(b1(c1(x1))) b0#(c1(c0(x1))) (64)
b1#(b1(b1(x1))) c1#(b0(x1)) (65)
b1#(b1(b1(x1))) c1#(b0(c1(b0(x1)))) (66)
b1#(b1(b1(x1))) b0#(x1) (67)
b1#(b1(b1(x1))) b0#(c1(b0(x1))) (68)

1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.