Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/88283)

The rewrite relation of the following TRS is considered.

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

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by matchbox @ termCOMP 2023)

1 Closure Under Flat Contexts

Using the flat contexts

{3(☐), 2(☐), 1(☐), 0(☐)}

We obtain the transformed TRS

There are 120 ruless (increase limit for explicit display).

1.1 Closure Under Flat Contexts

Using the flat contexts

{3(☐), 2(☐), 1(☐), 0(☐)}

We obtain the transformed TRS

There are 480 ruless (increase limit for explicit display).

1.1.1 Semantic Labeling

The following interpretations form a model of the rules.

As carrier we take the set {0,...,15}. Symbols are labeled by the interpretation of their arguments using the interpretations (modulo 16):

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

We obtain the labeled TRS

There are 7680 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
[30(x1)] = x1 +
0
[34(x1)] = x1 +
0
[38(x1)] = x1 +
297
[312(x1)] = x1 +
303
[31(x1)] = x1 +
243
[35(x1)] = x1 +
0
[39(x1)] = x1 +
297
[313(x1)] = x1 +
271
[32(x1)] = x1 +
297
[36(x1)] = x1 +
297
[310(x1)] = x1 +
297
[314(x1)] = x1 +
297
[33(x1)] = x1 +
298
[37(x1)] = x1 +
6
[311(x1)] = x1 +
298
[315(x1)] = x1 +
1
[20(x1)] = x1 +
0
[24(x1)] = x1 +
0
[28(x1)] = x1 +
0
[212(x1)] = x1 +
189
[21(x1)] = x1 +
0
[25(x1)] = x1 +
0
[29(x1)] = x1 +
0
[213(x1)] = x1 +
0
[22(x1)] = x1 +
0
[26(x1)] = x1 +
0
[210(x1)] = x1 +
0
[214(x1)] = x1 +
0
[23(x1)] = x1 +
297
[27(x1)] = x1 +
0
[211(x1)] = x1 +
297
[215(x1)] = x1 +
298
[10(x1)] = x1 +
243
[14(x1)] = x1 +
244
[18(x1)] = x1 +
297
[112(x1)] = x1 +
298
[11(x1)] = x1 +
0
[15(x1)] = x1 +
0
[19(x1)] = x1 +
297
[113(x1)] = x1 +
1
[12(x1)] = x1 +
297
[16(x1)] = x1 +
81
[110(x1)] = x1 +
297
[114(x1)] = x1 +
297
[13(x1)] = x1 +
297
[17(x1)] = x1 +
297
[111(x1)] = x1 +
303
[115(x1)] = x1 +
216
[00(x1)] = x1 +
276
[04(x1)] = x1 +
297
[08(x1)] = x1 +
297
[012(x1)] = x1 +
271
[01(x1)] = x1 +
297
[05(x1)] = x1 +
0
[09(x1)] = x1 +
1
[013(x1)] = x1 +
1
[02(x1)] = x1 +
298
[06(x1)] = x1 +
82
[010(x1)] = x1 +
297
[014(x1)] = x1 +
303
[03(x1)] = x1 +
303
[07(x1)] = x1 +
1
[011(x1)] = x1 +
298
[015(x1)] = x1 +
297
all of the following rules can be deleted.

There are 7680 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.