Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/86577)

The rewrite relation of the following TRS is considered.

0(0(1(1(x1)))) 0(2(1(x1))) (1)
0(0(3(3(0(0(x1)))))) 0(0(4(0(4(0(x1)))))) (2)
1(5(5(2(1(0(x1)))))) 1(5(4(3(1(0(x1)))))) (3)
4(0(3(4(2(2(x1)))))) 4(5(2(2(3(x1))))) (4)
2(2(4(1(3(4(1(x1))))))) 5(2(4(5(3(x1))))) (5)
0(3(1(5(2(3(1(5(x1)))))))) 0(3(3(5(3(5(1(5(x1)))))))) (6)
3(1(5(1(5(0(4(2(x1)))))))) 3(1(4(5(0(5(2(x1))))))) (7)
3(0(0(2(2(0(4(4(0(x1))))))))) 2(5(5(0(2(0(2(0(x1)))))))) (8)
3(0(1(3(1(5(1(4(2(x1))))))))) 1(1(2(4(2(4(1(0(2(x1))))))))) (9)
3(0(4(4(0(2(4(4(4(x1))))))))) 2(5(5(0(3(3(4(x1))))))) (10)
0(2(5(2(4(3(4(0(4(3(x1)))))))))) 0(3(5(3(1(3(4(0(3(x1))))))))) (11)
1(4(5(3(1(1(1(0(3(2(x1)))))))))) 1(3(2(5(3(5(2(1(x1)))))))) (12)
2(4(0(4(5(3(3(3(2(0(x1)))))))))) 0(3(1(5(4(5(5(0(x1)))))))) (13)
1(3(2(1(5(1(1(0(5(2(2(x1))))))))))) 1(0(2(5(4(2(2(0(5(3(x1)))))))))) (14)
3(2(4(0(1(0(0(2(0(1(3(x1))))))))))) 2(4(0(3(5(5(1(3(3(x1))))))))) (15)
4(4(1(0(1(3(3(1(4(1(2(x1))))))))))) 3(1(2(4(4(4(2(1(2(2(x1)))))))))) (16)
4(0(4(2(2(0(4(2(1(4(1(4(x1)))))))))))) 4(5(0(5(4(5(5(4(x1)))))))) (17)
4(3(5(4(3(4(0(0(4(2(5(0(x1)))))))))))) 4(0(2(0(3(3(3(5(3(1(2(4(x1)))))))))))) (18)
2(5(5(1(5(1(2(0(2(1(3(3(4(x1))))))))))))) 3(1(1(1(2(5(2(1(4(0(0(4(4(x1))))))))))))) (19)
3(4(1(1(3(1(1(1(5(4(3(4(1(x1))))))))))))) 3(4(0(0(3(5(3(5(0(4(4(4(x1)))))))))))) (20)
1(1(2(0(2(4(1(1(3(3(3(5(1(3(x1)))))))))))))) 5(1(3(4(5(1(1(0(3(4(0(2(0(x1))))))))))))) (21)
1(4(0(0(5(2(2(5(2(2(3(0(2(5(x1)))))))))))))) 1(5(4(2(3(0(1(3(0(3(2(2(0(5(x1)))))))))))))) (22)
2(2(0(2(0(0(5(3(2(3(2(0(3(2(x1)))))))))))))) 4(5(1(5(3(3(3(4(4(2(0(4(x1)))))))))))) (23)
4(2(3(5(4(2(5(5(1(1(4(4(0(4(1(0(x1)))))))))))))))) 4(1(2(5(4(0(5(3(2(5(0(4(2(4(0(x1))))))))))))))) (24)
5(0(3(4(0(0(0(4(3(4(2(4(3(3(2(0(4(x1))))))))))))))))) 5(2(3(3(1(2(0(4(3(0(1(5(5(2(3(4(x1)))))))))))))))) (25)
5(5(4(4(0(5(5(2(4(0(5(1(2(3(2(4(0(x1))))))))))))))))) 0(3(2(0(4(3(1(2(4(5(3(3(0(1(1(5(1(3(x1)))))))))))))))))) (26)
0(1(4(3(5(3(0(4(1(1(2(3(3(1(4(0(5(2(3(x1))))))))))))))))))) 0(3(0(0(5(4(5(0(4(0(2(2(1(4(0(0(0(4(x1)))))))))))))))))) (27)
4(2(1(0(0(4(4(4(3(0(5(1(3(1(2(3(5(1(0(2(x1)))))))))))))))))))) 4(4(1(4(5(0(4(2(4(5(2(3(4(2(2(5(5(1(x1)))))))))))))))))) (28)
4(4(4(1(4(5(2(2(0(1(4(5(2(2(1(4(5(0(0(4(x1)))))))))))))))))))) 2(1(2(1(4(0(4(5(3(2(5(0(3(2(0(2(2(5(1(4(x1)))))))))))))))))))) (29)
5(3(2(0(4(1(4(1(4(2(5(2(3(4(4(4(3(2(5(2(4(x1))))))))))))))))))))) 5(0(4(3(0(0(2(1(5(4(4(2(5(3(4(3(0(5(5(x1))))))))))))))))))) (30)

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(1(0(0(x1)))) 1(2(0(x1))) (31)
0(0(3(3(0(0(x1)))))) 0(4(0(4(0(0(x1)))))) (32)
0(1(2(5(5(1(x1)))))) 0(1(3(4(5(1(x1)))))) (33)
2(2(4(3(0(4(x1)))))) 3(2(2(5(4(x1))))) (34)
1(4(3(1(4(2(2(x1))))))) 3(5(4(2(5(x1))))) (35)
5(1(3(2(5(1(3(0(x1)))))))) 5(1(5(3(5(3(3(0(x1)))))))) (36)
2(4(0(5(1(5(1(3(x1)))))))) 2(5(0(5(4(1(3(x1))))))) (37)
0(4(4(0(2(2(0(0(3(x1))))))))) 0(2(0(2(0(5(5(2(x1)))))))) (38)
2(4(1(5(1(3(1(0(3(x1))))))))) 2(0(1(4(2(4(2(1(1(x1))))))))) (39)
4(4(4(2(0(4(4(0(3(x1))))))))) 4(3(3(0(5(5(2(x1))))))) (40)
3(4(0(4(3(4(2(5(2(0(x1)))))))))) 3(0(4(3(1(3(5(3(0(x1))))))))) (41)
2(3(0(1(1(1(3(5(4(1(x1)))))))))) 1(2(5(3(5(2(3(1(x1)))))))) (42)
0(2(3(3(3(5(4(0(4(2(x1)))))))))) 0(5(5(4(5(1(3(0(x1)))))))) (43)
2(2(5(0(1(1(5(1(2(3(1(x1))))))))))) 3(5(0(2(2(4(5(2(0(1(x1)))))))))) (44)
3(1(0(2(0(0(1(0(4(2(3(x1))))))))))) 3(3(1(5(5(3(0(4(2(x1))))))))) (45)
2(1(4(1(3(3(1(0(1(4(4(x1))))))))))) 2(2(1(2(4(4(4(2(1(3(x1)))))))))) (46)
4(1(4(1(2(4(0(2(2(4(0(4(x1)))))))))))) 4(5(5(4(5(0(5(4(x1)))))))) (47)
0(5(2(4(0(0(4(3(4(5(3(4(x1)))))))))))) 4(2(1(3(5(3(3(3(0(2(0(4(x1)))))))))))) (48)
4(3(3(1(2(0(2(1(5(1(5(5(2(x1))))))))))))) 4(4(0(0(4(1(2(5(2(1(1(1(3(x1))))))))))))) (49)
1(4(3(4(5(1(1(1(3(1(1(4(3(x1))))))))))))) 4(4(4(0(5(3(5(3(0(0(4(3(x1)))))))))))) (50)
3(1(5(3(3(3(1(1(4(2(0(2(1(1(x1)))))))))))))) 0(2(0(4(3(0(1(1(5(4(3(1(5(x1))))))))))))) (51)
5(2(0(3(2(2(5(2(2(5(0(0(4(1(x1)))))))))))))) 5(0(2(2(3(0(3(1(0(3(2(4(5(1(x1)))))))))))))) (52)
2(3(0(2(3(2(3(5(0(0(2(0(2(2(x1)))))))))))))) 4(0(2(4(4(3(3(3(5(1(5(4(x1)))))))))))) (53)
0(1(4(0(4(4(1(1(5(5(2(4(5(3(2(4(x1)))))))))))))))) 0(4(2(4(0(5(2(3(5(0(4(5(2(1(4(x1))))))))))))))) (54)
4(0(2(3(3(4(2(4(3(4(0(0(0(4(3(0(5(x1))))))))))))))))) 4(3(2(5(5(1(0(3(4(0(2(1(3(3(2(5(x1)))))))))))))))) (55)
0(4(2(3(2(1(5(0(4(2(5(5(0(4(4(5(5(x1))))))))))))))))) 3(1(5(1(1(0(3(3(5(4(2(1(3(4(0(2(3(0(x1)))))))))))))))))) (56)
3(2(5(0(4(1(3(3(2(1(1(4(0(3(5(3(4(1(0(x1))))))))))))))))))) 4(0(0(0(4(1(2(2(0(4(0(5(4(5(0(0(3(0(x1)))))))))))))))))) (57)
2(0(1(5(3(2(1(3(1(5(0(3(4(4(4(0(0(1(2(4(x1)))))))))))))))))))) 1(5(5(2(2(4(3(2(5(4(2(4(0(5(4(1(4(4(x1)))))))))))))))))) (58)
4(0(0(5(4(1(2(2(5(4(1(0(2(2(5(4(1(4(4(4(x1)))))))))))))))))))) 4(1(5(2(2(0(2(3(0(5(2(3(5(4(0(4(1(2(1(2(x1)))))))))))))))))))) (59)
4(2(5(2(3(4(4(4(3(2(5(2(4(1(4(1(4(0(2(3(5(x1))))))))))))))))))))) 5(5(0(3(4(3(5(2(4(4(5(1(2(0(0(3(4(0(5(x1))))))))))))))))))) (60)

1.1 Bounds

The given TRS is match-(raise)-bounded by 2. This is shown by the following automaton. The automaton is closed under rewriting as it is compatible.