Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/96438)

The rewrite relation of the following TRS is considered.

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