Certification Problem

Input (TPDB SRS_Standard/Secret_06_SRS/aprove08)

The rewrite relation of the following TRS is considered.

i(0(x1)) p(s(p(s(0(p(s(p(s(x1))))))))) (1)
i(s(x1)) p(s(p(s(s(j(p(s(p(s(p(p(p(p(s(s(s(s(x1)))))))))))))))))) (2)
j(0(x1)) p(s(p(p(s(s(0(p(s(p(s(x1))))))))))) (3)
j(s(x1)) s(s(s(s(p(p(s(s(i(p(s(p(s(x1))))))))))))) (4)
p(p(s(x1))) p(x1) (5)
p(s(x1)) x1 (6)
p(0(x1)) 0(s(s(s(s(s(s(s(s(x1))))))))) (7)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS
0(i(x1)) s(p(s(p(0(s(p(s(p(x1))))))))) (8)
s(i(x1)) s(s(s(s(p(p(p(p(s(p(s(p(j(s(s(p(s(p(x1)))))))))))))))))) (9)
0(j(x1)) s(p(s(p(0(s(s(p(p(s(p(x1))))))))))) (10)
s(j(x1)) s(p(s(p(i(s(s(p(p(s(s(s(s(x1))))))))))))) (11)
s(p(p(x1))) p(x1) (12)
s(p(x1)) x1 (13)
0(p(x1)) s(s(s(s(s(s(s(s(0(x1))))))))) (14)

1.1 Rule Removal

Using the linear polynomial interpretation over the arctic semiring over the integers
[j(x1)] = 3 · x1 + -∞
[i(x1)] = 3 · x1 + -∞
[p(x1)] = 0 · x1 + -∞
[0(x1)] = 8 · x1 + -∞
[s(x1)] = 0 · x1 + -∞
all of the following rules can be deleted.
0(i(x1)) s(p(s(p(0(s(p(s(p(x1))))))))) (8)
0(j(x1)) s(p(s(p(0(s(s(p(p(s(p(x1))))))))))) (10)

1.1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
s#(i(x1)) s#(p(x1)) (15)
s#(i(x1)) s#(p(s(p(x1)))) (16)
s#(i(x1)) s#(s(p(s(p(x1))))) (17)
s#(i(x1)) s#(p(j(s(s(p(s(p(x1)))))))) (18)
s#(i(x1)) s#(p(s(p(j(s(s(p(s(p(x1)))))))))) (19)
s#(i(x1)) s#(p(p(p(p(s(p(s(p(j(s(s(p(s(p(x1))))))))))))))) (20)
s#(i(x1)) s#(s(p(p(p(p(s(p(s(p(j(s(s(p(s(p(x1)))))))))))))))) (21)
s#(i(x1)) s#(s(s(p(p(p(p(s(p(s(p(j(s(s(p(s(p(x1))))))))))))))))) (22)
s#(i(x1)) s#(s(s(s(p(p(p(p(s(p(s(p(j(s(s(p(s(p(x1)))))))))))))))))) (23)
s#(j(x1)) s#(x1) (24)
s#(j(x1)) s#(s(x1)) (25)
s#(j(x1)) s#(s(s(x1))) (26)
s#(j(x1)) s#(s(s(s(x1)))) (27)
s#(j(x1)) s#(p(p(s(s(s(s(x1))))))) (28)
s#(j(x1)) s#(s(p(p(s(s(s(s(x1)))))))) (29)
s#(j(x1)) s#(p(i(s(s(p(p(s(s(s(s(x1))))))))))) (30)
s#(j(x1)) s#(p(s(p(i(s(s(p(p(s(s(s(s(x1))))))))))))) (31)
0#(p(x1)) 0#(x1) (32)
0#(p(x1)) s#(0(x1)) (33)
0#(p(x1)) s#(s(0(x1))) (34)
0#(p(x1)) s#(s(s(0(x1)))) (35)
0#(p(x1)) s#(s(s(s(0(x1))))) (36)
0#(p(x1)) s#(s(s(s(s(0(x1)))))) (37)
0#(p(x1)) s#(s(s(s(s(s(0(x1))))))) (38)
0#(p(x1)) s#(s(s(s(s(s(s(0(x1)))))))) (39)
0#(p(x1)) s#(s(s(s(s(s(s(s(0(x1))))))))) (40)

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.