Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/40093)

The rewrite relation of the following TRS is considered.

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

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by matchbox @ termCOMP 2023)

1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS
1(5(4(3(2(1(0(x1))))))) 5(4(3(2(1(0(5(4(3(2(1(0(1(1(5(4(3(2(1(x1))))))))))))))))))) (3)
1(5(4(3(2(1(0(x1))))))) 5(4(3(2(1(0(5(4(3(2(1(0(5(4(3(2(1(0(1(1(5(4(3(2(1(x1))))))))))))))))))))))))) (4)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
1#(5(4(3(2(1(0(x1))))))) 1#(x1) (5)
1#(5(4(3(2(1(0(x1))))))) 1#(5(4(3(2(1(x1)))))) (6)
1#(5(4(3(2(1(0(x1))))))) 1#(1(5(4(3(2(1(x1))))))) (7)
1#(5(4(3(2(1(0(x1))))))) 1#(0(5(4(3(2(1(0(5(4(3(2(1(0(1(1(5(4(3(2(1(x1))))))))))))))))))))) (8)
1#(5(4(3(2(1(0(x1))))))) 1#(0(5(4(3(2(1(0(1(1(5(4(3(2(1(x1))))))))))))))) (9)
1#(5(4(3(2(1(0(x1))))))) 1#(0(1(1(5(4(3(2(1(x1))))))))) (10)

1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.