Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/26103)

The rewrite relation of the following TRS is considered.

0(1(2(1(x1)))) → 1(2(1(1(0(1(2(0(1(2(x1)))))))))) (1)
0(1(2(1(x1)))) → 1(2(1(1(0(1(2(0(1(2(0(1(2(x1))))))))))))) (2)
0(1(2(1(x1)))) → 1(2(1(1(0(1(2(0(1(2(0(1(2(0(1(2(x1)))))))))))))))) (3)
0(1(2(1(x1)))) → 1(2(1(1(0(1(2(0(1(2(0(1(2(0(1(2(0(1(2(x1))))))))))))))))))) (4)
0(1(2(1(x1)))) → 1(2(1(1(0(1(2(0(1(2(0(1(2(0(1(2(0(1(2(0(1(2(x1)))))))))))))))))))))) (5)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by NaTT @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
0#(1(2(1(x1)))) → 0#(1(2(0(1(2(x1)))))) (6)
0#(1(2(1(x1)))) → 0#(1(2(0(1(2(0(1(2(0(1(2(x1)))))))))))) (7)
0#(1(2(1(x1)))) → 0#(1(2(x1))) (8)
0#(1(2(1(x1)))) → 0#(1(2(x1))) (8)
0#(1(2(1(x1)))) → 0#(1(2(0(1(2(0(1(2(x1))))))))) (9)
0#(1(2(1(x1)))) → 0#(1(2(0(1(2(x1)))))) (6)
0#(1(2(1(x1)))) → 0#(1(2(0(1(2(x1)))))) (6)
0#(1(2(1(x1)))) → 0#(1(2(x1))) (8)
0#(1(2(1(x1)))) → 0#(1(2(0(1(2(0(1(2(x1))))))))) (9)
0#(1(2(1(x1)))) → 0#(1(2(0(1(2(0(1(2(0(1(2(x1)))))))))))) (7)
0#(1(2(1(x1)))) → 0#(1(2(0(1(2(0(1(2(0(1(2(0(1(2(x1))))))))))))))) (10)
0#(1(2(1(x1)))) → 0#(1(2(0(1(2(0(1(2(0(1(2(x1)))))))))))) (7)
0#(1(2(1(x1)))) → 0#(1(2(0(1(2(x1)))))) (6)
0#(1(2(1(x1)))) → 0#(1(2(0(1(2(0(1(2(0(1(2(0(1(2(x1))))))))))))))) (10)
0#(1(2(1(x1)))) → 0#(1(2(x1))) (8)
0#(1(2(1(x1)))) → 0#(1(2(0(1(2(0(1(2(x1))))))))) (9)
0#(1(2(1(x1)))) → 0#(1(2(0(1(2(x1)))))) (6)
0#(1(2(1(x1)))) → 0#(1(2(x1))) (8)
0#(1(2(1(x1)))) → 0#(1(2(0(1(2(0(1(2(0(1(2(0(1(2(0(1(2(x1)))))))))))))))))) (11)
0#(1(2(1(x1)))) → 0#(1(2(0(1(2(0(1(2(x1))))))))) (9)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.