Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/57799)

The rewrite relation of the following TRS is considered.

There are 158 ruless (increase limit for explicit display).

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

There are 158 ruless (increase limit for explicit display).

1.1 Closure Under Flat Contexts

Using the flat contexts

{0(☐), 1(☐), 2(☐), 3(☐), 4(☐), 5(☐)}

We obtain the transformed TRS

There are 441 ruless (increase limit for explicit display).

1.1.1 Semantic Labeling

Root-labeling is applied.

We obtain the labeled TRS

There are 2646 ruless (increase limit for explicit display).

1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[01(x1)] = 1 · x1 + 4
[10(x1)] = 1 · x1 + 2
[00(x1)] = 1 · x1 + 2
[02(x1)] = 1 · x1
[21(x1)] = 1 · x1
[03(x1)] = 1 · x1
[05(x1)] = 1 · x1
[04(x1)] = 1 · x1 + 2
[31(x1)] = 1 · x1
[11(x1)] = 1 · x1 + 2
[12(x1)] = 1 · x1
[13(x1)] = 1 · x1
[15(x1)] = 1 · x1
[14(x1)] = 1 · x1
[54(x1)] = 1 · x1
[41(x1)] = 1 · x1
[20(x1)] = 1 · x1
[30(x1)] = 1 · x1 + 1
[40(x1)] = 1 · x1
[50(x1)] = 1 · x1
[22(x1)] = 1 · x1
[32(x1)] = 1 · x1
[52(x1)] = 1 · x1
[24(x1)] = 1 · x1
[34(x1)] = 1 · x1
[25(x1)] = 1 · x1
[51(x1)] = 1 · x1
[23(x1)] = 1 · x1
[33(x1)] = 1 · x1
[35(x1)] = 1 · x1
[42(x1)] = 1 · x1
[43(x1)] = 1 · x1 + 1
[45(x1)] = 1 · x1 + 2
[44(x1)] = 1 · x1
[53(x1)] = 1 · x1
[55(x1)] = 1 · x1
all of the following rules can be deleted.

There are 2568 ruless (increase limit for explicit display).

1.1.1.1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

There are 150 ruless (increase limit for explicit display).

1.1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.