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 AProVE @ termCOMP 2023)

1 Rule Removal

Using the linear polynomial interpretation over the naturals
[i(x1)] = 1 · x1 + 1
[0(x1)] = 1 · x1
[p(x1)] = 1 · x1
[s(x1)] = 1 · x1
[j(x1)] = 1 · x1 + 1
all of the following rules can be deleted.
i(0(x1)) p(s(p(s(0(p(s(p(s(x1))))))))) (1)
j(0(x1)) p(s(p(p(s(s(0(p(s(p(s(x1))))))))))) (3)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
i#(s(x1)) p#(s(p(s(s(j(p(s(p(s(p(p(p(p(s(s(s(s(x1)))))))))))))))))) (8)
i#(s(x1)) p#(s(s(j(p(s(p(s(p(p(p(p(s(s(s(s(x1)))))))))))))))) (9)
i#(s(x1)) j#(p(s(p(s(p(p(p(p(s(s(s(s(x1))))))))))))) (10)
i#(s(x1)) p#(s(p(s(p(p(p(p(s(s(s(s(x1)))))))))))) (11)
i#(s(x1)) p#(s(p(p(p(p(s(s(s(s(x1)))))))))) (12)
i#(s(x1)) p#(p(p(p(s(s(s(s(x1)))))))) (13)
i#(s(x1)) p#(p(p(s(s(s(s(x1))))))) (14)
i#(s(x1)) p#(p(s(s(s(s(x1)))))) (15)
i#(s(x1)) p#(s(s(s(s(x1))))) (16)
j#(s(x1)) p#(p(s(s(i(p(s(p(s(x1))))))))) (17)
j#(s(x1)) p#(s(s(i(p(s(p(s(x1)))))))) (18)
j#(s(x1)) i#(p(s(p(s(x1))))) (19)
j#(s(x1)) p#(s(p(s(x1)))) (20)
j#(s(x1)) p#(s(x1)) (21)
p#(p(s(x1))) p#(x1) (22)

1.1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.