Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/96256)

The rewrite relation of the following TRS is considered.

0(1(2(3(1(x1))))) 0(0(1(1(x1)))) (1)
4(5(2(3(5(x1))))) 4(4(0(5(x1)))) (2)
5(3(2(4(3(2(x1)))))) 5(3(0(5(4(x1))))) (3)
2(2(5(5(2(2(1(x1))))))) 3(2(3(3(1(0(x1)))))) (4)
4(5(2(1(1(3(4(x1))))))) 1(1(0(5(0(x1))))) (5)
0(2(2(3(3(2(5(0(4(x1))))))))) 0(2(2(1(3(4(1(4(x1)))))))) (6)
2(3(0(4(0(0(5(2(4(x1))))))))) 3(2(2(2(3(0(4(0(4(x1))))))))) (7)
2(4(3(3(3(5(0(1(3(x1))))))))) 0(1(0(0(5(0(5(x1))))))) (8)
2(3(5(3(3(4(1(5(1(2(x1)))))))))) 2(0(3(5(2(5(1(3(4(2(x1)))))))))) (9)
3(0(1(2(4(0(0(3(3(4(x1)))))))))) 3(1(5(0(4(2(0(2(5(4(x1)))))))))) (10)
1(1(4(0(4(2(1(2(4(5(3(x1))))))))))) 1(1(0(1(5(2(2(0(2(4(x1)))))))))) (11)
3(1(2(4(0(0(3(3(1(4(5(x1))))))))))) 2(3(1(1(4(0(4(2(2(2(5(x1))))))))))) (12)
4(3(0(4(1(0(4(1(5(0(3(x1))))))))))) 0(4(2(3(0(0(1(5(5(1(x1)))))))))) (13)
0(5(1(0(1(1(2(2(1(1(2(2(2(2(x1)))))))))))))) 0(3(0(5(3(1(3(0(3(4(2(3(3(1(4(x1))))))))))))))) (14)
0(5(2(1(1(4(2(5(2(0(1(3(3(3(x1)))))))))))))) 0(4(0(4(4(4(5(0(2(4(0(3(2(2(x1)))))))))))))) (15)
3(0(2(3(2(0(3(3(2(1(2(2(0(1(3(x1))))))))))))))) 2(1(5(3(2(1(2(2(0(4(0(1(3(1(x1)))))))))))))) (16)
0(5(2(3(4(0(5(3(2(4(2(2(4(0(3(4(x1)))))))))))))))) 0(0(0(5(1(3(2(2(3(0(1(3(2(1(x1)))))))))))))) (17)
4(4(2(5(3(3(1(1(3(1(1(0(5(3(1(4(x1)))))))))))))))) 4(2(4(4(4(2(1(5(5(4(3(3(0(5(1(5(4(x1))))))))))))))))) (18)
5(3(5(1(1(5(4(2(2(4(1(5(0(5(4(3(x1)))))))))))))))) 0(0(4(2(2(1(3(2(1(0(5(1(1(1(x1)))))))))))))) (19)
2(2(0(1(4(3(5(4(2(2(5(1(0(5(0(5(4(5(2(x1))))))))))))))))))) 3(3(4(2(0(4(5(2(5(2(0(0(4(4(3(5(1(4(5(2(x1)))))))))))))))))))) (20)
2(5(3(4(1(3(2(5(3(1(1(2(2(1(0(2(4(3(3(x1))))))))))))))))))) 0(3(5(2(3(1(1(0(3(4(3(4(0(0(5(2(2(3(x1)))))))))))))))))) (21)
5(3(1(5(0(1(4(4(2(4(3(2(5(0(4(1(0(4(4(x1))))))))))))))))))) 5(2(2(4(1(1(5(2(3(4(2(5(2(0(2(0(2(5(4(x1))))))))))))))))))) (22)
2(3(0(3(2(3(1(1(4(2(5(3(4(2(5(2(5(0(2(3(1(x1))))))))))))))))))))) 3(2(4(3(2(1(3(3(1(0(1(1(3(3(0(3(0(5(1(x1))))))))))))))))))) (23)
3(1(2(1(3(1(2(2(4(1(1(3(2(4(4(3(2(0(3(4(5(x1))))))))))))))))))))) 2(5(5(1(1(2(1(0(1(3(1(0(5(2(4(1(1(5(x1)))))))))))))))))) (24)
5(4(3(1(3(4(5(3(0(1(3(4(0(2(2(4(3(3(0(5(3(x1))))))))))))))))))))) 5(5(1(1(5(2(4(0(0(1(0(1(2(0(5(1(3(5(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
1(3(2(1(0(x1))))) 1(1(0(0(x1)))) (26)
5(3(2(5(4(x1))))) 5(0(4(4(x1)))) (27)
2(3(4(2(3(5(x1)))))) 4(5(0(3(5(x1))))) (28)
1(2(2(5(5(2(2(x1))))))) 0(1(3(3(2(3(x1)))))) (29)
4(3(1(1(2(5(4(x1))))))) 0(5(0(1(1(x1))))) (30)
4(0(5(2(3(3(2(2(0(x1))))))))) 4(1(4(3(1(2(2(0(x1)))))))) (31)
4(2(5(0(0(4(0(3(2(x1))))))))) 4(0(4(0(3(2(2(2(3(x1))))))))) (32)
3(1(0(5(3(3(3(4(2(x1))))))))) 5(0(5(0(0(1(0(x1))))))) (33)
2(1(5(1(4(3(3(5(3(2(x1)))))))))) 2(4(3(1(5(2(5(3(0(2(x1)))))))))) (34)
4(3(3(0(0(4(2(1(0(3(x1)))))))))) 4(5(2(0(2(4(0(5(1(3(x1)))))))))) (35)
3(5(4(2(1(2(4(0(4(1(1(x1))))))))))) 4(2(0(2(2(5(1(0(1(1(x1)))))))))) (36)
5(4(1(3(3(0(0(4(2(1(3(x1))))))))))) 5(2(2(2(4(0(4(1(1(3(2(x1))))))))))) (37)
3(0(5(1(4(0(1(4(0(3(4(x1))))))))))) 1(5(5(1(0(0(3(2(4(0(x1)))))))))) (38)
2(2(2(2(1(1(2(2(1(1(0(1(5(0(x1)))))))))))))) 4(1(3(3(2(4(3(0(3(1(3(5(0(3(0(x1))))))))))))))) (39)
3(3(3(1(0(2(5(2(4(1(1(2(5(0(x1)))))))))))))) 2(2(3(0(4(2(0(5(4(4(4(0(4(0(x1)))))))))))))) (40)
3(1(0(2(2(1(2(3(3(0(2(3(2(0(3(x1))))))))))))))) 1(3(1(0(4(0(2(2(1(2(3(5(1(2(x1)))))))))))))) (41)
4(3(0(4(2(2(4(2(3(5(0(4(3(2(5(0(x1)))))))))))))))) 1(2(3(1(0(3(2(2(3(1(5(0(0(0(x1)))))))))))))) (42)
4(1(3(5(0(1(1(3(1(1(3(3(5(2(4(4(x1)))))))))))))))) 4(5(1(5(0(3(3(4(5(5(1(2(4(4(4(2(4(x1))))))))))))))))) (43)
3(4(5(0(5(1(4(2(2(4(5(1(1(5(3(5(x1)))))))))))))))) 1(1(1(5(0(1(2(3(1(2(2(4(0(0(x1)))))))))))))) (44)
2(5(4(5(0(5(0(1(5(2(2(4(5(3(4(1(0(2(2(x1))))))))))))))))))) 2(5(4(1(5(3(4(4(0(0(2(5(2(5(4(0(2(4(3(3(x1)))))))))))))))))))) (45)
3(3(4(2(0(1(2(2(1(1(3(5(2(3(1(4(3(5(2(x1))))))))))))))))))) 3(2(2(5(0(0(4(3(4(3(0(1(1(3(2(5(3(0(x1)))))))))))))))))) (46)
4(4(0(1(4(0(5(2(3(4(2(4(4(1(0(5(1(3(5(x1))))))))))))))))))) 4(5(2(0(2(0(2(5(2(4(3(2(5(1(1(4(2(2(5(x1))))))))))))))))))) (47)
1(3(2(0(5(2(5(2(4(3(5(2(4(1(1(3(2(3(0(3(2(x1))))))))))))))))))))) 1(5(0(3(0(3(3(1(1(0(1(3(3(1(2(3(4(2(3(x1))))))))))))))))))) (48)
5(4(3(0(2(3(4(4(2(3(1(1(4(2(2(1(3(1(2(1(3(x1))))))))))))))))))))) 5(1(1(4(2(5(0(1(3(1(0(1(2(1(1(5(5(2(x1)))))))))))))))))) (49)
3(5(0(3(3(4(2(2(0(4(3(1(0(3(5(4(3(1(3(4(5(x1))))))))))))))))))))) 5(3(1(5(0(2(1(0(1(0(0(4(2(5(1(1(5(5(x1)))))))))))))))))) (50)

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.