Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/85874)

The rewrite relation of the following TRS is considered.

0(1(2(3(x1)))) 0(4(2(x1))) (1)
2(0(5(4(5(x1))))) 0(3(2(5(x1)))) (2)
1(2(3(2(3(1(x1)))))) 2(4(2(0(2(x1))))) (3)
3(5(2(1(2(2(x1)))))) 2(5(5(5(x1)))) (4)
5(1(0(4(1(3(x1)))))) 5(4(5(2(0(x1))))) (5)
0(3(1(4(0(1(2(x1))))))) 0(0(0(3(2(1(5(x1))))))) (6)
1(1(5(5(4(4(4(2(x1)))))))) 1(0(1(1(3(2(0(5(x1)))))))) (7)
2(0(2(0(3(5(1(5(x1)))))))) 2(5(1(0(5(0(1(0(x1)))))))) (8)
3(5(1(2(3(1(0(1(0(x1))))))))) 5(3(5(1(2(4(3(x1))))))) (9)
1(1(0(5(3(5(1(5(1(5(x1)))))))))) 1(5(0(2(0(2(2(1(1(3(3(x1))))))))))) (10)
2(2(1(2(4(2(0(5(0(4(x1)))))))))) 3(2(1(1(3(2(3(5(0(4(x1)))))))))) (11)
1(0(5(0(5(4(1(3(2(4(1(x1))))))))))) 3(5(4(3(4(5(3(0(5(x1))))))))) (12)
0(1(1(1(2(2(1(5(1(0(3(5(x1)))))))))))) 0(0(1(0(2(3(0(0(1(5(1(2(0(x1))))))))))))) (13)
2(5(4(3(5(3(3(2(4(0(4(3(x1)))))))))))) 5(3(2(3(1(0(5(4(0(4(2(5(x1)))))))))))) (14)
3(3(1(5(0(2(5(2(4(3(1(5(x1)))))))))))) 5(0(0(4(4(2(1(5(0(1(0(0(0(0(x1)))))))))))))) (15)
1(5(2(2(0(5(2(3(4(4(3(3(5(3(x1)))))))))))))) 2(5(2(1(0(3(1(2(3(4(5(4(4(0(x1)))))))))))))) (16)
3(2(3(3(0(2(3(5(4(2(3(3(2(3(x1)))))))))))))) 5(5(3(5(5(3(5(2(4(0(1(0(0(x1))))))))))))) (17)
3(4(3(2(5(1(1(5(5(3(0(0(2(0(x1)))))))))))))) 4(1(0(1(3(0(0(4(4(5(1(2(4(2(x1)))))))))))))) (18)
2(0(1(0(3(2(5(2(1(3(5(1(0(1(3(x1))))))))))))))) 0(0(5(3(2(3(1(3(1(3(2(0(3(2(x1)))))))))))))) (19)
0(2(5(2(0(5(3(2(4(3(5(1(4(4(4(5(x1)))))))))))))))) 2(4(4(5(4(4(1(3(2(3(1(4(5(2(2(5(x1)))))))))))))))) (20)
0(2(3(5(3(4(1(1(4(3(0(2(1(0(5(5(4(x1))))))))))))))))) 0(5(4(5(4(3(5(1(0(0(5(5(0(0(1(4(x1)))))))))))))))) (21)
4(4(2(2(1(1(1(1(2(3(0(3(5(1(4(1(3(x1))))))))))))))))) 4(5(2(0(3(3(2(1(1(3(0(0(5(3(3(3(x1)))))))))))))))) (22)
0(3(1(3(5(4(3(4(2(4(1(3(0(3(4(5(5(2(3(x1))))))))))))))))))) 0(3(2(4(3(3(4(4(5(5(3(3(1(2(4(2(2(0(x1)))))))))))))))))) (23)
3(3(0(1(0(4(1(4(2(1(0(5(3(3(2(1(3(5(0(x1))))))))))))))))))) 2(3(0(2(1(3(4(5(1(5(3(4(1(0(4(4(2(5(x1)))))))))))))))))) (24)
1(3(3(5(4(3(5(2(4(3(3(5(0(0(1(3(5(3(4(5(x1)))))))))))))))))))) 3(0(4(3(1(5(3(3(5(2(1(0(5(0(1(1(1(5(0(3(5(x1))))))))))))))))))))) (25)
2(0(5(1(5(0(1(0(4(4(5(2(4(2(5(3(4(3(0(5(x1)))))))))))))))))))) 2(3(2(2(1(5(4(0(0(2(3(4(0(1(4(4(2(5(5(x1))))))))))))))))))) (26)
3(0(4(2(5(1(5(3(3(2(4(3(0(0(5(1(0(5(4(3(x1)))))))))))))))))))) 3(2(1(4(2(0(4(2(4(4(2(5(1(3(3(4(3(4(0(x1))))))))))))))))))) (27)
3(2(0(1(0(5(1(3(1(4(3(5(4(5(5(5(5(5(1(5(x1)))))))))))))))))))) 3(1(4(4(5(3(5(2(1(4(4(0(1(3(4(1(3(1(5(3(5(x1))))))))))))))))))))) (28)
3(3(1(0(0(1(1(0(4(4(2(3(4(0(2(3(5(5(0(2(x1)))))))))))))))))))) 5(2(2(4(5(5(0(2(4(3(3(2(1(2(4(5(2(x1))))))))))))))))) (29)
3(0(5(5(1(5(5(1(1(3(5(0(1(1(3(1(5(4(4(4(5(x1))))))))))))))))))))) 5(2(4(3(1(4(3(3(2(3(0(4(0(1(2(1(4(2(2(0(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
3(2(1(0(x1)))) 2(4(0(x1))) (31)
5(4(5(0(2(x1))))) 5(2(3(0(x1)))) (32)
1(3(2(3(2(1(x1)))))) 2(0(2(4(2(x1))))) (33)
2(2(1(2(5(3(x1)))))) 5(5(5(2(x1)))) (34)
3(1(4(0(1(5(x1)))))) 0(2(5(4(5(x1))))) (35)
2(1(0(4(1(3(0(x1))))))) 5(1(2(3(0(0(0(x1))))))) (36)
2(4(4(4(5(5(1(1(x1)))))))) 5(0(2(3(1(1(0(1(x1)))))))) (37)
5(1(5(3(0(2(0(2(x1)))))))) 0(1(0(5(0(1(5(2(x1)))))))) (38)
0(1(0(1(3(2(1(5(3(x1))))))))) 3(4(2(1(5(3(5(x1))))))) (39)
5(1(5(1(5(3(5(0(1(1(x1)))))))))) 3(3(1(1(2(2(0(2(0(5(1(x1))))))))))) (40)
4(0(5(0(2(4(2(1(2(2(x1)))))))))) 4(0(5(3(2(3(1(1(2(3(x1)))))))))) (41)
1(4(2(3(1(4(5(0(5(0(1(x1))))))))))) 5(0(3(5(4(3(4(5(3(x1))))))))) (42)
5(3(0(1(5(1(2(2(1(1(1(0(x1)))))))))))) 0(2(1(5(1(0(0(3(2(0(1(0(0(x1))))))))))))) (43)
3(4(0(4(2(3(3(5(3(4(5(2(x1)))))))))))) 5(2(4(0(4(5(0(1(3(2(3(5(x1)))))))))))) (44)
5(1(3(4(2(5(2(0(5(1(3(3(x1)))))))))))) 0(0(0(0(1(0(5(1(2(4(4(0(0(5(x1)))))))))))))) (45)
3(5(3(3(4(4(3(2(5(0(2(2(5(1(x1)))))))))))))) 0(4(4(5(4(3(2(1(3(0(1(2(5(2(x1)))))))))))))) (46)
3(2(3(3(2(4(5(3(2(0(3(3(2(3(x1)))))))))))))) 0(0(1(0(4(2(5(3(5(5(3(5(5(x1))))))))))))) (47)
0(2(0(0(3(5(5(1(1(5(2(3(4(3(x1)))))))))))))) 2(4(2(1(5(4(4(0(0(3(1(0(1(4(x1)))))))))))))) (48)
3(1(0(1(5(3(1(2(5(2(3(0(1(0(2(x1))))))))))))))) 2(3(0(2(3(1(3(1(3(2(3(5(0(0(x1)))))))))))))) (49)
5(4(4(4(1(5(3(4(2(3(5(0(2(5(2(0(x1)))))))))))))))) 5(2(2(5(4(1(3(2(3(1(4(4(5(4(4(2(x1)))))))))))))))) (50)
4(5(5(0(1(2(0(3(4(1(1(4(3(5(3(2(0(x1))))))))))))))))) 4(1(0(0(5(5(0(0(1(5(3(4(5(4(5(0(x1)))))))))))))))) (51)
3(1(4(1(5(3(0(3(2(1(1(1(1(2(2(4(4(x1))))))))))))))))) 3(3(3(5(0(0(3(1(1(2(3(3(0(2(5(4(x1)))))))))))))))) (52)
3(2(5(5(4(3(0(3(1(4(2(4(3(4(5(3(1(3(0(x1))))))))))))))))))) 0(2(2(4(2(1(3(3(5(5(4(4(3(3(4(2(3(0(x1)))))))))))))))))) (53)
0(5(3(1(2(3(3(5(0(1(2(4(1(4(0(1(0(3(3(x1))))))))))))))))))) 5(2(4(4(0(1(4(3(5(1(5(4(3(1(2(0(3(2(x1)))))))))))))))))) (54)
5(4(3(5(3(1(0(0(5(3(3(4(2(5(3(4(5(3(3(1(x1)))))))))))))))))))) 5(3(0(5(1(1(1(0(5(0(1(2(5(3(3(5(1(3(4(0(3(x1))))))))))))))))))))) (55)
5(0(3(4(3(5(2(4(2(5(4(4(0(1(0(5(1(5(0(2(x1)))))))))))))))))))) 5(5(2(4(4(1(0(4(3(2(0(0(4(5(1(2(2(3(2(x1))))))))))))))))))) (56)
3(4(5(0(1(5(0(0(3(4(2(3(3(5(1(5(2(4(0(3(x1)))))))))))))))))))) 0(4(3(4(3(3(1(5(2(4(4(2(4(0(2(4(1(2(3(x1))))))))))))))))))) (57)
5(1(5(5(5(5(5(4(5(3(4(1(3(1(5(0(1(0(2(3(x1)))))))))))))))))))) 5(3(5(1(3(1(4(3(1(0(4(4(1(2(5(3(5(4(4(1(3(x1))))))))))))))))))))) (58)
2(0(5(5(3(2(0(4(3(2(4(4(0(1(1(0(0(1(3(3(x1)))))))))))))))))))) 2(5(4(2(1(2(3(3(4(2(0(5(5(4(2(2(5(x1))))))))))))))))) (59)
5(4(4(4(5(1(3(1(1(0(5(3(1(1(5(5(1(5(5(0(3(x1))))))))))))))))))))) 0(2(2(4(1(2(1(0(4(0(3(2(3(3(4(1(3(4(2(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.