Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/26291)

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)
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(0(1(2(x1))))))))))))))))))))))))) (6)
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(0(1(2(0(1(2(x1)))))))))))))))))))))))))))) (7)
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(0(1(2(0(1(2(0(1(2(x1))))))))))))))))))))))))))))))) (8)
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(0(1(2(0(1(2(0(1(2(0(1(2(x1)))))))))))))))))))))))))))))))))) (9)
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(0(1(2(0(1(2(0(1(2(0(1(2(0(1(2(x1))))))))))))))))))))))))))))))))))))) (10)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS
1(2(1(0(x1)))) 2(1(0(2(1(0(1(1(2(1(x1)))))))))) (11)
1(2(1(0(x1)))) 2(1(0(2(1(0(2(1(0(1(1(2(1(x1))))))))))))) (12)
1(2(1(0(x1)))) 2(1(0(2(1(0(2(1(0(2(1(0(1(1(2(1(x1)))))))))))))))) (13)
1(2(1(0(x1)))) 2(1(0(2(1(0(2(1(0(2(1(0(2(1(0(1(1(2(1(x1))))))))))))))))))) (14)
1(2(1(0(x1)))) 2(1(0(2(1(0(2(1(0(2(1(0(2(1(0(2(1(0(1(1(2(1(x1)))))))))))))))))))))) (15)
1(2(1(0(x1)))) 2(1(0(2(1(0(2(1(0(2(1(0(2(1(0(2(1(0(2(1(0(1(1(2(1(x1))))))))))))))))))))))))) (16)
1(2(1(0(x1)))) 2(1(0(2(1(0(2(1(0(2(1(0(2(1(0(2(1(0(2(1(0(2(1(0(1(1(2(1(x1)))))))))))))))))))))))))))) (17)
1(2(1(0(x1)))) 2(1(0(2(1(0(2(1(0(2(1(0(2(1(0(2(1(0(2(1(0(2(1(0(2(1(0(1(1(2(1(x1))))))))))))))))))))))))))))))) (18)
1(2(1(0(x1)))) 2(1(0(2(1(0(2(1(0(2(1(0(2(1(0(2(1(0(2(1(0(2(1(0(2(1(0(2(1(0(1(1(2(1(x1)))))))))))))))))))))))))))))))))) (19)
1(2(1(0(x1)))) 2(1(0(2(1(0(2(1(0(2(1(0(2(1(0(2(1(0(2(1(0(2(1(0(2(1(0(2(1(0(2(1(0(1(1(2(1(x1))))))))))))))))))))))))))))))))))))) (20)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
1#(2(1(0(x1)))) 1#(x1) (21)
1#(2(1(0(x1)))) 1#(2(1(x1))) (22)
1#(2(1(0(x1)))) 1#(1(2(1(x1)))) (23)
1#(2(1(0(x1)))) 1#(0(1(1(2(1(x1)))))) (24)
1#(2(1(0(x1)))) 1#(0(2(1(0(1(1(2(1(x1))))))))) (25)
1#(2(1(0(x1)))) 1#(0(2(1(0(2(1(0(1(1(2(1(x1)))))))))))) (26)
1#(2(1(0(x1)))) 1#(0(2(1(0(2(1(0(2(1(0(1(1(2(1(x1))))))))))))))) (27)
1#(2(1(0(x1)))) 1#(0(2(1(0(2(1(0(2(1(0(2(1(0(1(1(2(1(x1)))))))))))))))))) (28)
1#(2(1(0(x1)))) 1#(0(2(1(0(2(1(0(2(1(0(2(1(0(2(1(0(1(1(2(1(x1))))))))))))))))))))) (29)
1#(2(1(0(x1)))) 1#(0(2(1(0(2(1(0(2(1(0(2(1(0(2(1(0(2(1(0(1(1(2(1(x1)))))))))))))))))))))))) (30)
1#(2(1(0(x1)))) 1#(0(2(1(0(2(1(0(2(1(0(2(1(0(2(1(0(2(1(0(2(1(0(1(1(2(1(x1))))))))))))))))))))))))))) (31)
1#(2(1(0(x1)))) 1#(0(2(1(0(2(1(0(2(1(0(2(1(0(2(1(0(2(1(0(2(1(0(2(1(0(1(1(2(1(x1)))))))))))))))))))))))))))))) (32)
1#(2(1(0(x1)))) 1#(0(2(1(0(2(1(0(2(1(0(2(1(0(2(1(0(2(1(0(2(1(0(2(1(0(2(1(0(1(1(2(1(x1))))))))))))))))))))))))))))))))) (33)
1#(2(1(0(x1)))) 1#(0(2(1(0(2(1(0(2(1(0(2(1(0(2(1(0(2(1(0(2(1(0(2(1(0(2(1(0(2(1(0(1(1(2(1(x1)))))))))))))))))))))))))))))))))))) (34)

1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.