Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/48267)

The rewrite relation of the following TRS is considered.

0(1(2(x1))) → 0(0(2(1(x1)))) (1)
3(0(1(x1))) → 0(0(3(1(x1)))) (2)
0(1(4(3(x1)))) → 0(0(4(1(3(x1))))) (3)
0(1(4(4(x1)))) → 0(0(4(4(1(x1))))) (4)
0(4(2(1(x1)))) → 0(2(0(4(1(x1))))) (5)
2(1(0(1(x1)))) → 0(0(2(1(1(x1))))) (6)
2(2(0(1(x1)))) → 2(1(0(0(2(x1))))) (7)
3(0(4(3(x1)))) → 3(0(0(0(3(4(x1)))))) (8)
4(0(1(3(x1)))) → 1(0(0(3(4(x1))))) (9)
0(4(0(1(2(x1))))) → 0(2(4(1(0(0(0(0(x1)))))))) (10)
0(5(4(0(5(x1))))) → 0(4(5(0(0(0(0(0(5(x1))))))))) (11)
2(0(1(5(1(x1))))) → 0(0(2(5(1(1(x1)))))) (12)
2(0(5(3(5(x1))))) → 0(3(5(0(2(5(x1)))))) (13)
2(2(0(5(2(x1))))) → 0(2(5(0(2(0(2(x1))))))) (14)
2(2(3(0(1(x1))))) → 0(0(2(2(1(3(x1)))))) (15)
2(5(0(1(4(x1))))) → 2(0(0(5(4(1(x1)))))) (16)
2(5(4(0(1(x1))))) → 0(2(0(4(5(1(x1)))))) (17)
5(0(4(2(3(x1))))) → 5(0(0(2(4(3(x1)))))) (18)
5(1(0(1(2(x1))))) → 0(2(0(5(1(1(x1)))))) (19)
0(4(1(2(0(1(x1)))))) → 0(0(1(4(0(2(1(x1))))))) (20)
0(4(2(2(0(1(x1)))))) → 0(2(2(1(0(0(4(x1))))))) (21)
0(4(3(3(5(1(x1)))))) → 0(0(4(3(1(5(3(x1))))))) (22)
2(4(5(3(0(1(x1)))))) → 0(3(0(2(4(5(1(x1))))))) (23)
2(5(3(0(4(2(x1)))))) → 0(0(2(4(5(3(2(x1))))))) (24)
3(0(1(4(2(2(x1)))))) → 3(4(1(0(2(0(2(x1))))))) (25)
3(2(1(4(0(3(x1)))))) → 0(0(2(3(3(4(1(x1))))))) (26)
3(5(3(0(1(3(x1)))))) → 3(5(3(1(0(0(3(x1))))))) (27)
4(3(2(2(4(4(x1)))))) → 4(3(4(0(2(2(4(x1))))))) (28)
4(3(4(5(1(3(x1)))))) → 4(3(1(4(5(0(3(x1))))))) (29)
5(0(1(2(5(2(x1)))))) → 5(2(0(1(5(0(0(2(x1)))))))) (30)
5(0(4(2(3(1(x1)))))) → 5(3(4(1(0(0(2(x1))))))) (31)
5(1(1(3(0(4(x1)))))) → 0(0(1(5(3(1(4(x1))))))) (32)
5(1(4(2(0(5(x1)))))) → 0(2(4(1(5(0(5(x1))))))) (33)
5(4(0(1(2(2(x1)))))) → 1(5(0(2(4(0(2(x1))))))) (34)
0(0(5(3(2(1(2(x1))))))) → 0(0(5(3(1(0(2(2(x1)))))))) (35)
0(1(0(1(4(5(1(x1))))))) → 0(0(0(1(1(4(5(3(1(x1))))))))) (36)
0(1(2(5(3(0(1(x1))))))) → 0(0(0(1(5(1(2(3(x1)))))))) (37)
0(1(3(0(1(2(1(x1))))))) → 0(0(1(2(1(5(3(1(x1)))))))) (38)
0(1(3(0(5(4(1(x1))))))) → 0(5(1(3(4(0(0(0(1(x1))))))))) (39)
0(1(5(3(0(3(2(x1))))))) → 0(0(2(3(3(1(4(5(x1)))))))) (40)
0(4(0(3(5(2(5(x1))))))) → 0(2(0(0(0(3(4(5(5(x1))))))))) (41)
0(4(2(0(4(3(2(x1))))))) → 2(4(0(0(2(0(3(4(x1)))))))) (42)
0(4(2(1(0(5(2(x1))))))) → 4(1(0(0(0(2(5(2(x1)))))))) (43)
0(4(2(2(5(5(2(x1))))))) → 2(5(5(4(0(2(0(2(x1)))))))) (44)
2(0(1(2(4(4(2(x1))))))) → 2(0(0(0(1(4(4(2(2(x1))))))))) (45)
2(1(2(0(3(0(5(x1))))))) → 0(0(3(2(5(0(2(1(x1)))))))) (46)
2(3(0(4(3(0(5(x1))))))) → 0(3(3(5(2(0(0(4(x1)))))))) (47)
2(4(5(3(3(5(5(x1))))))) → 0(2(5(5(5(3(3(4(x1)))))))) (48)
2(5(5(2(4(0(5(x1))))))) → 2(0(0(4(5(2(5(5(x1)))))))) (49)
3(0(0(3(1(5(1(x1))))))) → 3(1(5(0(0(0(3(1(x1)))))))) (50)
3(0(0(3(2(2(2(x1))))))) → 3(3(2(0(2(0(0(2(x1)))))))) (51)
3(0(4(2(0(4(2(x1))))))) → 3(0(2(0(4(4(2(0(x1)))))))) (52)
3(0(5(5(5(3(2(x1))))))) → 3(2(5(5(0(0(3(5(x1)))))))) (53)
4(3(3(3(2(1(2(x1))))))) → 1(0(3(3(3(4(2(2(x1)))))))) (54)
4(4(3(0(1(1(1(x1))))))) → 4(0(0(4(3(1(1(1(x1)))))))) (55)
5(1(0(3(0(5(2(x1))))))) → 0(2(0(5(3(3(1(5(x1)))))))) (56)
5(1(2(0(0(5(1(x1))))))) → 5(1(0(0(0(2(5(1(x1)))))))) (57)
5(4(0(1(1(0(5(x1))))))) → 1(5(0(0(0(5(4(1(x1)))))))) (58)
5(4(3(2(1(2(3(x1))))))) → 5(2(0(2(3(3(1(4(x1)))))))) (59)
0(1(3(3(5(2(1(4(x1)))))))) → 0(5(1(3(3(4(1(0(2(x1))))))))) (60)
0(4(2(2(0(1(0(4(x1)))))))) → 0(2(0(0(4(1(0(2(4(x1))))))))) (61)
0(4(2(2(5(0(5(2(x1)))))))) → 0(2(2(0(0(2(4(5(5(x1))))))))) (62)
2(1(5(0(1(3(2(4(x1)))))))) → 2(5(0(2(1(3(3(1(4(x1))))))))) (63)
2(4(3(4(1(2(3(2(x1)))))))) → 2(1(3(4(3(4(2(0(2(x1))))))))) (64)
2(4(5(3(0(1(1(1(x1)))))))) → 0(0(2(1(1(5(3(4(1(x1))))))))) (65)
2(5(1(2(4(0(5(2(x1)))))))) → 0(4(1(5(2(5(2(0(0(2(x1)))))))))) (66)
4(0(1(2(4(5(3(2(x1)))))))) → 4(0(2(4(3(1(4(5(2(x1))))))))) (67)
5(0(5(2(3(4(0(1(x1)))))))) → 5(5(4(0(3(0(0(2(1(x1))))))))) (68)
5(1(5(1(2(0(3(5(x1)))))))) → 0(0(2(5(3(5(1(1(5(x1))))))))) (69)
5(4(0(3(1(1(2(4(x1)))))))) → 2(3(1(5(0(0(0(4(1(4(x1)))))))))) (70)
5(5(3(2(1(1(0(4(x1)))))))) → 4(1(5(0(2(5(3(3(1(x1))))))))) (71)
0(3(3(0(4(0(2(2(5(x1))))))))) → 3(5(0(2(4(3(0(0(0(0(2(x1))))))))))) (72)
0(4(0(0(1(0(1(4(2(x1))))))))) → 0(4(1(0(0(2(0(2(4(1(x1)))))))))) (73)
0(5(0(1(3(0(3(4(5(x1))))))))) → 5(0(3(3(4(0(0(0(1(5(x1)))))))))) (74)
2(1(2(4(0(5(1(3(5(x1))))))))) → 0(0(4(2(2(1(1(5(3(5(x1)))))))))) (75)
2(3(0(5(2(5(0(4(2(x1))))))))) → 2(0(5(3(4(2(0(2(5(2(x1)))))))))) (76)
2(3(4(2(0(1(2(1(2(x1))))))))) → 2(0(2(1(2(1(3(4(0(2(x1)))))))))) (77)
2(4(1(3(0(3(0(4(1(x1))))))))) → 0(0(0(3(4(3(2(4(1(1(x1)))))))))) (78)
3(0(1(0(3(3(0(5(1(x1))))))))) → 3(0(1(0(0(0(5(3(3(1(x1)))))))))) (79)
4(1(1(3(0(5(2(5(4(x1))))))))) → 1(3(5(0(0(2(4(5(1(4(x1)))))))))) (80)
5(2(0(4(0(4(2(3(1(x1))))))))) → 5(0(0(3(1(4(4(2(0(2(x1)))))))))) (81)
5(3(0(3(0(1(3(4(3(x1))))))))) → 5(3(4(3(1(0(0(0(3(4(3(x1))))))))))) (82)
0(1(0(1(3(2(5(0(1(4(x1)))))))))) → 0(2(3(0(0(3(1(5(4(1(1(x1))))))))))) (83)
2(1(1(0(2(3(1(0(5(5(x1)))))))))) → 0(2(2(3(1(5(1(1(5(0(0(x1))))))))))) (84)
2(5(0(4(0(5(1(0(3(3(x1)))))))))) → 2(5(4(5(1(3(0(0(0(0(3(x1))))))))))) (85)
2(5(1(1(2(5(0(4(3(2(x1)))))))))) → 2(5(1(5(2(4(3(0(0(2(1(x1))))))))))) (86)
3(2(0(1(2(0(4(4(2(5(x1)))))))))) → 3(0(0(2(5(1(2(3(4(4(2(x1))))))))))) (87)
4(1(4(2(5(0(1(1(0(1(x1)))))))))) → 4(5(4(1(0(0(1(0(2(1(1(x1))))))))))) (88)
4(2(1(4(0(1(1(3(0(3(x1)))))))))) → 1(4(0(0(2(1(3(4(1(3(4(x1))))))))))) (89)
5(4(3(3(5(5(5(1(0(2(x1)))))))))) → 5(3(3(4(0(2(5(0(5(5(1(x1))))))))))) (90)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by matchbox @ termCOMP 2023)

1 Closure Under Flat Contexts

Using the flat contexts

{5(☐), 4(☐), 3(☐), 2(☐), 1(☐), 0(☐)}

We obtain the transformed TRS

There are 540 ruless (increase limit for explicit display).

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 3240 ruless (increase limit for explicit display).

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 +
165
[51(x1)] = x1 +
165
[52(x1)] = x1 +
77
[53(x1)] = x1 +
165
[54(x1)] = x1 +
132
[55(x1)] = x1 +
1
[40(x1)] = x1 +
155
[41(x1)] = x1 +
81
[42(x1)] = x1 +
154
[43(x1)] = x1 +
169
[44(x1)] = x1 +
0
[45(x1)] = x1 +
165
[30(x1)] = x1 +
166
[31(x1)] = x1 +
0
[32(x1)] = x1 +
4
[33(x1)] = x1 +
169
[34(x1)] = x1 +
0
[35(x1)] = x1 +
165
[20(x1)] = x1 +
88
[21(x1)] = x1 +
100
[22(x1)] = x1 +
275/2
[23(x1)] = x1 +
165
[24(x1)] = x1 +
33
[25(x1)] = x1 +
0
[10(x1)] = x1 +
0
[11(x1)] = x1 +
100
[12(x1)] = x1 +
4
[13(x1)] = x1 +
169
[14(x1)] = x1 +
22
[15(x1)] = x1 +
143
[00(x1)] = x1 +
165
[01(x1)] = x1 +
169
[02(x1)] = x1 +
33
[03(x1)] = x1 +
0
[04(x1)] = x1 +
166
[05(x1)] = x1 +
0
all of the following rules can be deleted.

There are 3240 ruless (increase limit for explicit display).

1.1.1.1 R is empty

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