Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/186023)

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 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS

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

1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[3(x1)] = 1 · x1 + 6
[1(x1)] = 1 · x1 + 3
[5(x1)] = 1 · x1
[0(x1)] = 1 · x1 + 4
[2(x1)] = 1 · x1 + 5
[4(x1)] = 1 · x1 + 1
all of the following rules can be deleted.

There are 179 ruless (increase limit for explicit display).

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.
4#(4(2(4(1(3(3(0(0(2(4(5(0(3(4(0(x1)))))))))))))))) → 4#(4(1(0(0(4(5(4(2(3(2(0(3(3(4(0(x1)))))))))))))))) (361)
4#(4(2(4(1(3(3(0(0(2(4(5(0(3(4(0(x1)))))))))))))))) → 4#(1(0(0(4(5(4(2(3(2(0(3(3(4(0(x1))))))))))))))) (362)
4#(4(2(4(1(3(3(0(0(2(4(5(0(3(4(0(x1)))))))))))))))) → 4#(5(4(2(3(2(0(3(3(4(0(x1))))))))))) (363)
4#(4(2(4(1(3(3(0(0(2(4(5(0(3(4(0(x1)))))))))))))))) → 4#(2(3(2(0(3(3(4(0(x1))))))))) (364)

1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.