Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/3633)

The rewrite relation of the following TRS is considered.

0(4(4(x1))) 1(2(2(5(1(2(2(2(3(4(x1)))))))))) (1)
5(3(5(4(x1)))) 5(0(4(0(2(3(2(2(5(4(x1)))))))))) (2)
0(1(4(4(1(x1))))) 1(4(2(3(0(5(3(3(4(1(x1)))))))))) (3)
0(3(2(1(3(x1))))) 0(1(2(3(1(5(1(5(1(5(x1)))))))))) (4)
0(3(5(4(1(x1))))) 0(5(5(3(4(4(2(5(0(2(x1)))))))))) (5)
0(4(1(0(1(x1))))) 3(2(2(2(3(0(5(1(4(1(x1)))))))))) (6)
1(0(4(4(1(x1))))) 0(5(3(3(3(1(5(4(3(0(x1)))))))))) (7)
3(5(5(5(5(x1))))) 0(5(3(3(1(5(0(1(1(5(x1)))))))))) (8)
4(0(0(0(4(x1))))) 4(2(2(4(2(4(2(5(2(4(x1)))))))))) (9)
4(0(0(1(3(x1))))) 4(2(3(1(4(2(2(5(1(2(x1)))))))))) (10)
4(1(0(5(5(x1))))) 4(2(3(2(0(2(2(3(0(5(x1)))))))))) (11)
5(0(1(0(0(x1))))) 5(0(2(2(0(1(2(2(2(2(x1)))))))))) (12)
5(5(5(0(4(x1))))) 3(3(1(4(2(0(5(1(2(4(x1)))))))))) (13)
0(1(3(5(5(1(x1)))))) 4(3(0(5(0(3(4(2(5(1(x1)))))))))) (14)
0(3(1(0(5(3(x1)))))) 2(3(0(5(2(3(5(0(5(3(x1)))))))))) (15)
1(0(3(2(1(4(x1)))))) 2(4(3(2(3(1(1(2(5(4(x1)))))))))) (16)
1(1(0(3(1(0(x1)))))) 2(0(2(4(0(3(3(0(2(5(x1)))))))))) (17)
1(1(4(1(3(0(x1)))))) 2(0(5(4(5(3(3(0(5(0(x1)))))))))) (18)
1(3(1(1(4(5(x1)))))) 1(3(5(0(2(2(5(1(5(5(x1)))))))))) (19)
1(5(4(1(0(3(x1)))))) 3(3(3(3(4(0(5(4(0(5(x1)))))))))) (20)
2(1(0(1(0(2(x1)))))) 5(1(5(1(5(4(3(4(1(2(x1)))))))))) (21)
3(3(3(2(0(0(x1)))))) 3(3(4(2(2(5(2(5(2(2(x1)))))))))) (22)
3(5(5(1(4(1(x1)))))) 0(5(4(3(5(3(1(5(1(5(x1)))))))))) (23)
4(5(5(5(3(2(x1)))))) 2(4(2(4(2(3(5(2(3(2(x1)))))))))) (24)
5(1(3(5(1(0(x1)))))) 2(5(2(3(1(0(2(2(2(5(x1)))))))))) (25)
5(3(2(5(5(0(x1)))))) 5(1(2(2(2(4(1(1(4(2(x1)))))))))) (26)
0(0(3(2(0(0(0(x1))))))) 3(4(4(1(2(2(2(2(2(0(x1)))))))))) (27)
0(1(1(3(5(2(5(x1))))))) 0(0(3(2(2(4(5(4(3(5(x1)))))))))) (28)
0(1(4(0(3(5(5(x1))))))) 1(3(4(0(5(3(2(1(0(5(x1)))))))))) (29)
0(4(4(0(3(5(1(x1))))))) 0(4(2(3(2(3(5(1(2(1(x1)))))))))) (30)
0(4(4(5(2(4(3(x1))))))) 2(5(2(3(1(2(5(2(4(3(x1)))))))))) (31)
0(5(3(5(1(4(5(x1))))))) 1(3(1(3(4(2(3(2(5(3(x1)))))))))) (32)
1(0(5(0(1(3(2(x1))))))) 2(2(0(5(3(3(4(2(3(2(x1)))))))))) (33)
1(1(0(0(4(1(2(x1))))))) 2(0(1(2(2(4(2(2(3(2(x1)))))))))) (34)
1(1(0(1(3(5(5(x1))))))) 0(5(3(2(3(1(1(2(2(5(x1)))))))))) (35)
1(1(0(3(1(0(0(x1))))))) 5(2(3(1(1(2(5(4(2(2(x1)))))))))) (36)
2(1(0(0(3(5(2(x1))))))) 2(0(4(2(5(0(2(1(4(2(x1)))))))))) (37)
2(4(1(5(0(3(5(x1))))))) 2(2(1(2(5(3(0(2(4(5(x1)))))))))) (38)
3(0(0(4(1(3(1(x1))))))) 4(3(5(4(0(5(1(2(4(0(x1)))))))))) (39)
3(1(4(0(1(4(1(x1))))))) 4(2(5(0(2(1(5(0(4(2(x1)))))))))) (40)
3(5(0(1(0(4(3(x1))))))) 3(5(4(3(2(3(0(4(2(3(x1)))))))))) (41)
3(5(0(4(3(4(4(x1))))))) 3(5(5(0(2(3(3(1(5(4(x1)))))))))) (42)
3(5(1(3(5(4(1(x1))))))) 3(2(3(4(0(2(1(5(4(3(x1)))))))))) (43)
4(1(5(4(1(1(1(x1))))))) 2(3(2(4(2(5(1(1(4(1(x1)))))))))) (44)
5(1(3(5(1(0(0(x1))))))) 5(3(4(1(1(3(3(0(2(0(x1)))))))))) (45)
5(3(3(2(1(0(2(x1))))))) 5(4(2(1(3(3(1(5(4(2(x1)))))))))) (46)
5(4(1(1(1(4(5(x1))))))) 2(5(1(5(3(2(1(3(1(5(x1)))))))))) (47)
5(5(5(2(5(5(3(x1))))))) 2(5(4(2(0(0(5(0(5(3(x1)))))))))) (48)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

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.