Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/96563)

The rewrite relation of the following TRS is considered.

0(1(2(3(x1)))) 0(2(2(3(x1)))) (1)
1(1(4(4(5(x1))))) 5(1(0(0(5(x1))))) (2)
2(3(3(2(5(x1))))) 3(5(4(5(5(x1))))) (3)
5(1(1(5(3(x1))))) 5(1(4(5(3(x1))))) (4)
0(2(5(1(1(4(x1)))))) 0(3(2(2(4(5(x1)))))) (5)
1(2(1(4(0(3(x1)))))) 4(2(5(4(5(3(x1)))))) (6)
1(5(2(0(4(3(x1)))))) 5(0(1(1(1(x1))))) (7)
2(2(1(5(4(0(x1)))))) 2(3(5(4(0(3(x1)))))) (8)
4(0(2(3(1(4(4(x1))))))) 4(1(3(0(5(0(5(x1))))))) (9)
4(3(5(4(3(2(2(2(x1)))))))) 4(1(2(3(4(1(4(x1))))))) (10)
0(2(2(2(3(4(3(3(3(x1))))))))) 0(5(4(0(5(3(5(5(3(x1))))))))) (11)
4(5(3(4(1(1(0(5(2(x1))))))))) 4(4(3(1(3(2(0(5(2(x1))))))))) (12)
3(3(4(1(0(4(5(1(5(2(x1)))))))))) 3(3(4(1(5(0(3(0(2(4(x1)))))))))) (13)
2(4(5(4(0(5(5(1(2(1(4(2(x1)))))))))))) 4(1(0(1(1(3(3(0(1(4(2(x1))))))))))) (14)
1(0(0(4(5(4(3(4(5(1(5(4(0(x1))))))))))))) 5(3(3(1(2(1(1(0(3(1(3(x1))))))))))) (15)
1(2(2(1(2(3(2(4(4(0(0(3(2(x1))))))))))))) 4(4(4(3(2(0(1(3(1(2(2(4(5(x1))))))))))))) (16)
4(1(5(4(5(3(3(3(1(2(0(2(0(x1))))))))))))) 4(1(4(3(4(4(0(1(3(1(1(5(x1)))))))))))) (17)
2(2(3(3(0(2(0(2(5(0(5(5(5(1(0(3(1(x1))))))))))))))))) 5(5(5(2(0(2(5(0(0(4(3(2(1(4(3(3(1(x1))))))))))))))))) (18)
1(5(1(1(3(0(3(3(3(3(1(1(5(1(5(2(5(2(x1)))))))))))))))))) 3(5(4(2(0(5(1(3(4(1(1(2(5(2(4(1(5(2(x1)))))))))))))))))) (19)
2(4(3(3(4(2(3(5(4(0(4(0(2(3(0(3(3(0(x1)))))))))))))))))) 2(1(3(0(4(4(1(5(2(1(0(3(0(5(0(1(5(x1))))))))))))))))) (20)
2(5(5(1(2(3(5(0(3(0(0(3(3(1(0(1(3(2(x1)))))))))))))))))) 2(1(5(0(0(1(5(2(2(2(3(1(0(5(0(4(3(3(x1)))))))))))))))))) (21)
2(3(5(1(3(0(0(5(0(2(2(4(5(4(0(3(1(4(2(x1))))))))))))))))))) 5(2(0(2(0(3(2(3(5(2(2(4(1(3(5(5(1(2(x1)))))))))))))))))) (22)
3(2(3(4(1(2(5(5(0(1(2(3(4(0(5(1(4(4(5(4(x1)))))))))))))))))))) 3(2(0(1(5(0(1(5(5(1(2(3(0(2(4(2(5(2(2(4(x1)))))))))))))))))))) (23)
3(5(5(2(0(4(4(5(5(0(3(5(2(1(2(1(0(2(2(3(x1)))))))))))))))))))) 3(1(3(3(0(5(4(2(4(1(2(3(5(0(3(4(5(2(4(3(x1)))))))))))))))))))) (24)
4(5(1(4(5(3(1(5(5(3(5(0(2(1(2(4(4(5(1(0(4(x1))))))))))))))))))))) 4(4(0(2(1(5(2(0(4(5(2(2(3(5(5(2(2(0(3(5(4(x1))))))))))))))))))))) (25)

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
3(2(1(0(x1)))) 3(2(2(0(x1)))) (26)
5(4(4(1(1(x1))))) 5(0(0(1(5(x1))))) (27)
5(2(3(3(2(x1))))) 5(5(4(5(3(x1))))) (28)
3(5(1(1(5(x1))))) 3(5(4(1(5(x1))))) (29)
4(1(1(5(2(0(x1)))))) 5(4(2(2(3(0(x1)))))) (30)
3(0(4(1(2(1(x1)))))) 3(5(4(5(2(4(x1)))))) (31)
3(4(0(2(5(1(x1)))))) 1(1(1(0(5(x1))))) (32)
0(4(5(1(2(2(x1)))))) 3(0(4(5(3(2(x1)))))) (33)
4(4(1(3(2(0(4(x1))))))) 5(0(5(0(3(1(4(x1))))))) (34)
2(2(2(3(4(5(3(4(x1)))))))) 4(1(4(3(2(1(4(x1))))))) (35)
3(3(3(4(3(2(2(2(0(x1))))))))) 3(5(5(3(5(0(4(5(0(x1))))))))) (36)
2(5(0(1(1(4(3(5(4(x1))))))))) 2(5(0(2(3(1(3(4(4(x1))))))))) (37)
2(5(1(5(4(0(1(4(3(3(x1)))))))))) 4(2(0(3(0(5(1(4(3(3(x1)))))))))) (38)
2(4(1(2(1(5(5(0(4(5(4(2(x1)))))))))))) 2(4(1(0(3(3(1(1(0(1(4(x1))))))))))) (39)
0(4(5(1(5(4(3(4(5(4(0(0(1(x1))))))))))))) 3(1(3(0(1(1(2(1(3(3(5(x1))))))))))) (40)
2(3(0(0(4(4(2(3(2(1(2(2(1(x1))))))))))))) 5(4(2(2(1(3(1(0(2(3(4(4(4(x1))))))))))))) (41)
0(2(0(2(1(3(3(3(5(4(5(1(4(x1))))))))))))) 5(1(1(3(1(0(4(4(3(4(1(4(x1)))))))))))) (42)
1(3(0(1(5(5(5(0(5(2(0(2(0(3(3(2(2(x1))))))))))))))))) 1(3(3(4(1(2(3(4(0(0(5(2(0(2(5(5(5(x1))))))))))))))))) (43)
2(5(2(5(1(5(1(1(3(3(3(3(0(3(1(1(5(1(x1)))))))))))))))))) 2(5(1(4(2(5(2(1(1(4(3(1(5(0(2(4(5(3(x1)))))))))))))))))) (44)
0(3(3(0(3(2(0(4(0(4(5(3(2(4(3(3(4(2(x1)))))))))))))))))) 5(1(0(5(0(3(0(1(2(5(1(4(4(0(3(1(2(x1))))))))))))))))) (45)
2(3(1(0(1(3(3(0(0(3(0(5(3(2(1(5(5(2(x1)))))))))))))))))) 3(3(4(0(5(0(1(3(2(2(2(5(1(0(0(5(1(2(x1)))))))))))))))))) (46)
2(4(1(3(0(4(5(4(2(2(0(5(0(0(3(1(5(3(2(x1))))))))))))))))))) 2(1(5(5(3(1(4(2(2(5(3(2(3(0(2(0(2(5(x1)))))))))))))))))) (47)
4(5(4(4(1(5(0(4(3(2(1(0(5(5(2(1(4(3(2(3(x1)))))))))))))))))))) 4(2(2(5(2(4(2(0(3(2(1(5(5(1(0(5(1(0(2(3(x1)))))))))))))))))))) (48)
3(2(2(0(1(2(1(2(5(3(0(5(5(4(4(0(2(5(5(3(x1)))))))))))))))))))) 3(4(2(5(4(3(0(5(3(2(1(4(2(4(5(0(3(3(1(3(x1)))))))))))))))))))) (49)
4(0(1(5(4(4(2(1(2(0(5(3(5(5(1(3(5(4(1(5(4(x1))))))))))))))))))))) 4(5(3(0(2(2(5(5(3(2(2(5(4(0(2(5(1(2(0(4(4(x1))))))))))))))))))))) (50)

1.1 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.