Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/64160)

The rewrite relation of the following TRS is considered.

1(2(1(1(0(0(1(0(2(1(1(0(2(x1))))))))))))) 1(2(2(1(0(0(2(0(1(2(2(0(1(0(0(2(2(x1))))))))))))))))) (1)
2(0(1(2(1(2(1(2(1(0(0(1(0(x1))))))))))))) 2(2(0(0(2(1(1(1(0(0(0(1(1(1(2(2(0(x1))))))))))))))))) (2)
2(1(0(0(2(1(1(1(0(1(2(2(0(x1))))))))))))) 0(0(2(1(0(0(1(1(2(2(2(2(2(1(2(1(0(x1))))))))))))))))) (3)
2(2(0(0(1(0(2(1(1(1(0(0(1(x1))))))))))))) 2(2(1(1(2(0(0(1(2(2(0(2(0(0(2(2(2(x1))))))))))))))))) (4)
2(2(1(0(1(0(1(1(0(0(0(2(0(x1))))))))))))) 0(2(2(1(0(0(1(0(0(1(1(1(1(0(0(2(0(x1))))))))))))))))) (5)
2(2(1(1(0(2(0(0(0(2(2(0(0(x1))))))))))))) 0(0(1(2(0(0(0(2(2(0(1(2(2(2(2(0(0(x1))))))))))))))))) (6)
2(2(2(1(0(2(0(0(1(0(1(0(2(x1))))))))))))) 0(2(2(0(0(2(2(2(1(2(0(0(2(2(0(2(2(x1))))))))))))))))) (7)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 Bounds

The given TRS is match-(raise)-bounded by 1. This is shown by the following automaton. The automaton is closed under rewriting as it is compatible.