Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/96417)

The rewrite relation of the following TRS is considered.

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