Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/3746)

The rewrite relation of the following TRS is considered.

0(5(0(x1))) 3(4(3(3(2(0(4(4(4(0(x1)))))))))) (1)
0(2(5(0(x1)))) 0(2(1(1(5(2(1(4(1(3(x1)))))))))) (2)
0(5(0(5(x1)))) 3(2(0(2(0(4(4(1(5(4(x1)))))))))) (3)
2(5(5(3(x1)))) 4(3(0(2(1(0(1(3(4(3(x1)))))))))) (4)
5(5(5(0(x1)))) 4(1(2(1(2(3(5(0(1(3(x1)))))))))) (5)
0(5(1(5(0(x1))))) 3(4(0(1(4(5(2(2(3(1(x1)))))))))) (6)
3(0(0(5(3(x1))))) 1(3(4(3(5(2(4(1(3(3(x1)))))))))) (7)
3(5(5(0(0(x1))))) 1(4(1(0(0(4(4(0(4(1(x1)))))))))) (8)
0(2(5(1(5(0(x1)))))) 2(1(0(1(5(2(4(0(2(0(x1)))))))))) (9)
0(2(5(5(1(4(x1)))))) 4(0(1(1(1(1(0(4(1(5(x1)))))))))) (10)
0(5(3(1(2(5(x1)))))) 0(2(3(1(1(2(4(4(5(5(x1)))))))))) (11)
0(5(5(1(5(1(x1)))))) 0(5(1(1(3(3(4(2(1(0(x1)))))))))) (12)
4(5(5(4(2(0(x1)))))) 2(4(0(1(3(4(4(4(1(0(x1)))))))))) (13)
5(0(0(3(5(2(x1)))))) 4(4(2(3(0(1(2(0(5(2(x1)))))))))) (14)
5(1(5(0(2(5(x1)))))) 5(0(0(1(4(2(3(2(1(5(x1)))))))))) (15)
5(2(0(2(5(5(x1)))))) 5(5(4(4(4(5(4(4(1(4(x1)))))))))) (16)
5(5(0(2(5(0(x1)))))) 2(0(5(0(2(1(0(0(3(0(x1)))))))))) (17)
5(5(0(3(4(5(x1)))))) 2(0(5(5(2(1(3(2(3(2(x1)))))))))) (18)
5(5(3(5(0(5(x1)))))) 5(4(4(3(5(1(3(3(4(5(x1)))))))))) (19)
0(4(4(0(0(5(1(x1))))))) 1(3(2(0(4(1(5(1(1(2(x1)))))))))) (20)
0(4(4(2(5(5(5(x1))))))) 0(2(4(5(5(4(2(0(1(1(x1)))))))))) (21)
1(0(2(5(2(0(0(x1))))))) 3(1(4(4(0(3(0(1(2(2(x1)))))))))) (22)
1(2(0(4(2(5(0(x1))))))) 4(2(4(0(3(2(2(4(1(0(x1)))))))))) (23)
1(2(5(5(0(3(3(x1))))))) 3(4(1(2(0(3(3(1(0(3(x1)))))))))) (24)
1(5(5(3(3(3(4(x1))))))) 1(5(1(0(0(2(2(2(3(5(x1)))))))))) (25)
2(5(4(5(2(5(1(x1))))))) 4(3(2(1(4(2(2(4(5(2(x1)))))))))) (26)
3(2(3(5(1(5(2(x1))))))) 2(0(3(2(3(2(1(5(5(1(x1)))))))))) (27)
3(3(4(2(5(5(2(x1))))))) 1(2(3(3(4(4(1(4(0(1(x1)))))))))) (28)
3(5(0(5(5(5(0(x1))))))) 0(0(3(0(3(5(0(3(2(0(x1)))))))))) (29)
4(3(1(2(5(2(4(x1))))))) 2(3(1(1(4(3(4(4(2(4(x1)))))))))) (30)
4(5(5(3(1(0(5(x1))))))) 1(1(5(2(0(3(3(3(2(1(x1)))))))))) (31)
5(0(5(3(1(0(5(x1))))))) 5(2(4(4(2(1(3(5(1(5(x1)))))))))) (32)
5(0(5(3(5(1(5(x1))))))) 5(1(1(2(4(0(0(3(2(5(x1)))))))))) (33)
5(1(5(3(3(0(5(x1))))))) 4(4(3(2(2(2(5(0(1(1(x1)))))))))) (34)
5(2(0(2(5(3(3(x1))))))) 5(2(0(5(1(1(3(2(0(3(x1)))))))))) (35)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by matchbox @ termCOMP 2023)

1 Rule Removal

Using the matrix interpretations of dimension 1 with strict dimension 1 over the rationals with delta = 1
[5(x1)] = x1 +
1
[4(x1)] = x1 +
0
[3(x1)] = x1 +
0
[2(x1)] = x1 +
0
[1(x1)] = x1 +
0
[0(x1)] = x1 +
0
all of the following rules can be deleted.
0(5(0(x1))) 3(4(3(3(2(0(4(4(4(0(x1)))))))))) (1)
0(5(0(5(x1)))) 3(2(0(2(0(4(4(1(5(4(x1)))))))))) (3)
2(5(5(3(x1)))) 4(3(0(2(1(0(1(3(4(3(x1)))))))))) (4)
5(5(5(0(x1)))) 4(1(2(1(2(3(5(0(1(3(x1)))))))))) (5)
0(5(1(5(0(x1))))) 3(4(0(1(4(5(2(2(3(1(x1)))))))))) (6)
3(5(5(0(0(x1))))) 1(4(1(0(0(4(4(0(4(1(x1)))))))))) (8)
0(2(5(1(5(0(x1)))))) 2(1(0(1(5(2(4(0(2(0(x1)))))))))) (9)
0(2(5(5(1(4(x1)))))) 4(0(1(1(1(1(0(4(1(5(x1)))))))))) (10)
0(5(5(1(5(1(x1)))))) 0(5(1(1(3(3(4(2(1(0(x1)))))))))) (12)
4(5(5(4(2(0(x1)))))) 2(4(0(1(3(4(4(4(1(0(x1)))))))))) (13)
5(0(0(3(5(2(x1)))))) 4(4(2(3(0(1(2(0(5(2(x1)))))))))) (14)
5(1(5(0(2(5(x1)))))) 5(0(0(1(4(2(3(2(1(5(x1)))))))))) (15)
5(5(0(2(5(0(x1)))))) 2(0(5(0(2(1(0(0(3(0(x1)))))))))) (17)
5(5(0(3(4(5(x1)))))) 2(0(5(5(2(1(3(2(3(2(x1)))))))))) (18)
5(5(3(5(0(5(x1)))))) 5(4(4(3(5(1(3(3(4(5(x1)))))))))) (19)
0(4(4(2(5(5(5(x1))))))) 0(2(4(5(5(4(2(0(1(1(x1)))))))))) (21)
1(0(2(5(2(0(0(x1))))))) 3(1(4(4(0(3(0(1(2(2(x1)))))))))) (22)
1(2(0(4(2(5(0(x1))))))) 4(2(4(0(3(2(2(4(1(0(x1)))))))))) (23)
1(2(5(5(0(3(3(x1))))))) 3(4(1(2(0(3(3(1(0(3(x1)))))))))) (24)
2(5(4(5(2(5(1(x1))))))) 4(3(2(1(4(2(2(4(5(2(x1)))))))))) (26)
3(3(4(2(5(5(2(x1))))))) 1(2(3(3(4(4(1(4(0(1(x1)))))))))) (28)
3(5(0(5(5(5(0(x1))))))) 0(0(3(0(3(5(0(3(2(0(x1)))))))))) (29)
4(3(1(2(5(2(4(x1))))))) 2(3(1(1(4(3(4(4(2(4(x1)))))))))) (30)
4(5(5(3(1(0(5(x1))))))) 1(1(5(2(0(3(3(3(2(1(x1)))))))))) (31)
5(0(5(3(5(1(5(x1))))))) 5(1(1(2(4(0(0(3(2(5(x1)))))))))) (33)
5(1(5(3(3(0(5(x1))))))) 4(4(3(2(2(2(5(0(1(1(x1)))))))))) (34)

1.1 Closure Under Flat Contexts

Using the flat contexts

{5(), 4(), 3(), 2(), 1(), 0()}

We obtain the transformed TRS
5(0(2(5(0(x1))))) 5(0(2(1(1(5(2(1(4(1(3(x1))))))))))) (36)
5(3(0(0(5(3(x1)))))) 5(1(3(4(3(5(2(4(1(3(3(x1))))))))))) (37)
5(0(5(3(1(2(5(x1))))))) 5(0(2(3(1(1(2(4(4(5(5(x1))))))))))) (38)
5(5(2(0(2(5(5(x1))))))) 5(5(5(4(4(4(5(4(4(1(4(x1))))))))))) (39)
5(0(4(4(0(0(5(1(x1)))))))) 5(1(3(2(0(4(1(5(1(1(2(x1))))))))))) (40)
5(1(5(5(3(3(3(4(x1)))))))) 5(1(5(1(0(0(2(2(2(3(5(x1))))))))))) (41)
5(3(2(3(5(1(5(2(x1)))))))) 5(2(0(3(2(3(2(1(5(5(1(x1))))))))))) (42)
5(5(0(5(3(1(0(5(x1)))))))) 5(5(2(4(4(2(1(3(5(1(5(x1))))))))))) (43)
5(5(2(0(2(5(3(3(x1)))))))) 5(5(2(0(5(1(1(3(2(0(3(x1))))))))))) (44)
4(0(2(5(0(x1))))) 4(0(2(1(1(5(2(1(4(1(3(x1))))))))))) (45)
4(3(0(0(5(3(x1)))))) 4(1(3(4(3(5(2(4(1(3(3(x1))))))))))) (46)
4(0(5(3(1(2(5(x1))))))) 4(0(2(3(1(1(2(4(4(5(5(x1))))))))))) (47)
4(5(2(0(2(5(5(x1))))))) 4(5(5(4(4(4(5(4(4(1(4(x1))))))))))) (48)
4(0(4(4(0(0(5(1(x1)))))))) 4(1(3(2(0(4(1(5(1(1(2(x1))))))))))) (49)
4(1(5(5(3(3(3(4(x1)))))))) 4(1(5(1(0(0(2(2(2(3(5(x1))))))))))) (50)
4(3(2(3(5(1(5(2(x1)))))))) 4(2(0(3(2(3(2(1(5(5(1(x1))))))))))) (51)
4(5(0(5(3(1(0(5(x1)))))))) 4(5(2(4(4(2(1(3(5(1(5(x1))))))))))) (52)
4(5(2(0(2(5(3(3(x1)))))))) 4(5(2(0(5(1(1(3(2(0(3(x1))))))))))) (53)
3(0(2(5(0(x1))))) 3(0(2(1(1(5(2(1(4(1(3(x1))))))))))) (54)
3(3(0(0(5(3(x1)))))) 3(1(3(4(3(5(2(4(1(3(3(x1))))))))))) (55)
3(0(5(3(1(2(5(x1))))))) 3(0(2(3(1(1(2(4(4(5(5(x1))))))))))) (56)
3(5(2(0(2(5(5(x1))))))) 3(5(5(4(4(4(5(4(4(1(4(x1))))))))))) (57)
3(0(4(4(0(0(5(1(x1)))))))) 3(1(3(2(0(4(1(5(1(1(2(x1))))))))))) (58)
3(1(5(5(3(3(3(4(x1)))))))) 3(1(5(1(0(0(2(2(2(3(5(x1))))))))))) (59)
3(3(2(3(5(1(5(2(x1)))))))) 3(2(0(3(2(3(2(1(5(5(1(x1))))))))))) (60)
3(5(0(5(3(1(0(5(x1)))))))) 3(5(2(4(4(2(1(3(5(1(5(x1))))))))))) (61)
3(5(2(0(2(5(3(3(x1)))))))) 3(5(2(0(5(1(1(3(2(0(3(x1))))))))))) (62)
2(0(2(5(0(x1))))) 2(0(2(1(1(5(2(1(4(1(3(x1))))))))))) (63)
2(3(0(0(5(3(x1)))))) 2(1(3(4(3(5(2(4(1(3(3(x1))))))))))) (64)
2(0(5(3(1(2(5(x1))))))) 2(0(2(3(1(1(2(4(4(5(5(x1))))))))))) (65)
2(5(2(0(2(5(5(x1))))))) 2(5(5(4(4(4(5(4(4(1(4(x1))))))))))) (66)
2(0(4(4(0(0(5(1(x1)))))))) 2(1(3(2(0(4(1(5(1(1(2(x1))))))))))) (67)
2(1(5(5(3(3(3(4(x1)))))))) 2(1(5(1(0(0(2(2(2(3(5(x1))))))))))) (68)
2(3(2(3(5(1(5(2(x1)))))))) 2(2(0(3(2(3(2(1(5(5(1(x1))))))))))) (69)
2(5(0(5(3(1(0(5(x1)))))))) 2(5(2(4(4(2(1(3(5(1(5(x1))))))))))) (70)
2(5(2(0(2(5(3(3(x1)))))))) 2(5(2(0(5(1(1(3(2(0(3(x1))))))))))) (71)
1(0(2(5(0(x1))))) 1(0(2(1(1(5(2(1(4(1(3(x1))))))))))) (72)
1(3(0(0(5(3(x1)))))) 1(1(3(4(3(5(2(4(1(3(3(x1))))))))))) (73)
1(0(5(3(1(2(5(x1))))))) 1(0(2(3(1(1(2(4(4(5(5(x1))))))))))) (74)
1(5(2(0(2(5(5(x1))))))) 1(5(5(4(4(4(5(4(4(1(4(x1))))))))))) (75)
1(0(4(4(0(0(5(1(x1)))))))) 1(1(3(2(0(4(1(5(1(1(2(x1))))))))))) (76)
1(1(5(5(3(3(3(4(x1)))))))) 1(1(5(1(0(0(2(2(2(3(5(x1))))))))))) (77)
1(3(2(3(5(1(5(2(x1)))))))) 1(2(0(3(2(3(2(1(5(5(1(x1))))))))))) (78)
1(5(0(5(3(1(0(5(x1)))))))) 1(5(2(4(4(2(1(3(5(1(5(x1))))))))))) (79)
1(5(2(0(2(5(3(3(x1)))))))) 1(5(2(0(5(1(1(3(2(0(3(x1))))))))))) (80)
0(0(2(5(0(x1))))) 0(0(2(1(1(5(2(1(4(1(3(x1))))))))))) (81)
0(3(0(0(5(3(x1)))))) 0(1(3(4(3(5(2(4(1(3(3(x1))))))))))) (82)
0(0(5(3(1(2(5(x1))))))) 0(0(2(3(1(1(2(4(4(5(5(x1))))))))))) (83)
0(5(2(0(2(5(5(x1))))))) 0(5(5(4(4(4(5(4(4(1(4(x1))))))))))) (84)
0(0(4(4(0(0(5(1(x1)))))))) 0(1(3(2(0(4(1(5(1(1(2(x1))))))))))) (85)
0(1(5(5(3(3(3(4(x1)))))))) 0(1(5(1(0(0(2(2(2(3(5(x1))))))))))) (86)
0(3(2(3(5(1(5(2(x1)))))))) 0(2(0(3(2(3(2(1(5(5(1(x1))))))))))) (87)
0(5(0(5(3(1(0(5(x1)))))))) 0(5(2(4(4(2(1(3(5(1(5(x1))))))))))) (88)
0(5(2(0(2(5(3(3(x1)))))))) 0(5(2(0(5(1(1(3(2(0(3(x1))))))))))) (89)

1.1.1 Semantic Labeling

The following interpretations form a model of the rules.

As carrier we take the set {0,...,5}. Symbols are labeled by the interpretation of their arguments using the interpretations (modulo 6):

[5(x1)] = 6x1 + 0
[4(x1)] = 6x1 + 1
[3(x1)] = 6x1 + 2
[2(x1)] = 6x1 + 3
[1(x1)] = 6x1 + 4
[0(x1)] = 6x1 + 5

We obtain the labeled TRS

There are 324 ruless (increase limit for explicit display).

1.1.1.1 Rule Removal

Using the matrix interpretations of dimension 1 with strict dimension 1 over the rationals with delta = 1
[50(x1)] = x1 +
0
[51(x1)] = x1 +
0
[52(x1)] = x1 +
9
[53(x1)] = x1 +
1
[54(x1)] = x1 +
1
[55(x1)] = x1 +
3
[40(x1)] = x1 +
0
[41(x1)] = x1 +
0
[42(x1)] = x1 +
3
[43(x1)] = x1 +
0
[44(x1)] = x1 +
0
[45(x1)] = x1 +
0
[30(x1)] = x1 +
1
[31(x1)] = x1 +
3
[32(x1)] = x1 +
1
[33(x1)] = x1 +
0
[34(x1)] = x1 +
0
[35(x1)] = x1 +
0
[20(x1)] = x1 +
1
[21(x1)] = x1 +
1
[22(x1)] = x1 +
1
[23(x1)] = x1 +
0
[24(x1)] = x1 +
0
[25(x1)] = x1 +
0
[10(x1)] = x1 +
0
[11(x1)] = x1 +
0
[12(x1)] = x1 +
0
[13(x1)] = x1 +
0
[14(x1)] = x1 +
0
[15(x1)] = x1 +
0
[00(x1)] = x1 +
3
[01(x1)] = x1 +
9
[02(x1)] = x1 +
0
[03(x1)] = x1 +
0
[04(x1)] = x1 +
0
[05(x1)] = x1 +
0
all of the following rules can be deleted.

There are 324 ruless (increase limit for explicit display).

1.1.1.1.1 R is empty

There are no rules in the TRS. Hence, it is terminating.