Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/187254)

The rewrite relation of the following TRS is considered.

There are 180 ruless (increase limit for explicit display).

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
[0(x1)] = 1 · x1 + 16
[1(x1)] = 1 · x1 + 27
[2(x1)] = 1 · x1 + 12
[3(x1)] = 1 · x1 + 20
[4(x1)] = 1 · x1 + 11
[5(x1)] = 1 · x1
all of the following rules can be deleted.

There are 177 ruless (increase limit for explicit display).

1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[0(x1)] = 1 · x1
[2(x1)] = 1 · x1 + 1
[4(x1)] = 1 · x1
[3(x1)] = 1 · x1
[1(x1)] = 1 · x1
[5(x1)] = 1 · x1
all of the following rules can be deleted.
0(3(5(0(5(0(3(2(1(5(1(2(0(5(5(4(x1)))))))))))))))) → 0(2(4(3(4(1(3(5(5(5(5(4(1(4(0(4(x1)))))))))))))))) (83)
0(3(5(0(5(0(3(2(1(5(1(2(0(5(5(4(x1)))))))))))))))) → 0(4(4(1(5(5(5(4(1(3(2(0(4(3(5(4(x1)))))))))))))))) (84)

1.1.1 Switch to Innermost Termination

The TRS is overlay and locally confluent:

10

Hence, it suffices to show innermost termination in the following.

1.1.1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
0#(2(4(3(4(1(3(5(5(5(5(4(1(4(0(4(x1)))))))))))))))) → 0#(4(4(1(5(5(5(4(1(3(2(0(4(3(5(4(x1)))))))))))))))) (181)
0#(2(4(3(4(1(3(5(5(5(5(4(1(4(0(4(x1)))))))))))))))) → 0#(4(3(5(4(x1))))) (182)

1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.