Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/57132)

The rewrite relation of the following TRS is considered.

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

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 240 ruless (increase limit for explicit display).

1.1 Closure Under Flat Contexts

Using the flat contexts

{5(☐), 4(☐), 3(☐), 2(☐), 1(☐), 0(☐)}

We obtain the transformed TRS

There are 1440 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,...,35}. Symbols are labeled by the interpretation of their arguments using the interpretations (modulo 36):

[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 51840 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 +
2223
[56(x1)] = x1 +
2223
[512(x1)] = x1 +
348
[518(x1)] = x1 +
2223
[524(x1)] = x1 +
2223
[530(x1)] = x1 +
2223
[51(x1)] = x1 +
2224
[57(x1)] = x1 +
2223
[513(x1)] = x1 +
1
[519(x1)] = x1 +
1773
[525(x1)] = x1 +
2224
[531(x1)] = x1 +
2223
[52(x1)] = x1 +
1
[58(x1)] = x1 +
1
[514(x1)] = x1 +
0
[520(x1)] = x1 +
1767
[526(x1)] = x1 +
861
[532(x1)] = x1 +
2223
[53(x1)] = x1 +
2229
[59(x1)] = x1 +
1369
[515(x1)] = x1 +
2223
[521(x1)] = x1 +
1710
[527(x1)] = x1 +
2223
[533(x1)] = x1 +
1
[54(x1)] = x1 +
2229
[510(x1)] = x1 +
2223
[516(x1)] = x1 +
0
[522(x1)] = x1 +
2223
[528(x1)] = x1 +
2223
[534(x1)] = x1 +
2223
[55(x1)] = x1 +
2229
[511(x1)] = x1 +
2223
[517(x1)] = x1 +
2223
[523(x1)] = x1 +
2223
[529(x1)] = x1 +
2229
[535(x1)] = x1 +
2223
[40(x1)] = x1 +
2229
[46(x1)] = x1 +
2223
[412(x1)] = x1 +
0
[418(x1)] = x1 +
1
[424(x1)] = x1 +
2229
[430(x1)] = x1 +
2223
[41(x1)] = x1 +
2229
[47(x1)] = x1 +
0
[413(x1)] = x1 +
1881
[419(x1)] = x1 +
2223
[425(x1)] = x1 +
0
[431(x1)] = x1 +
6
[42(x1)] = x1 +
0
[48(x1)] = x1 +
0
[414(x1)] = x1 +
0
[420(x1)] = x1 +
1
[426(x1)] = x1 +
2223
[432(x1)] = x1 +
405
[43(x1)] = x1 +
6
[49(x1)] = x1 +
2223
[415(x1)] = x1 +
2223
[421(x1)] = x1 +
2223
[427(x1)] = x1 +
6
[433(x1)] = x1 +
399
[44(x1)] = x1 +
2223
[410(x1)] = x1 +
6
[416(x1)] = x1 +
1
[422(x1)] = x1 +
6
[428(x1)] = x1 +
0
[434(x1)] = x1 +
2223
[45(x1)] = x1 +
2224
[411(x1)] = x1 +
6
[417(x1)] = x1 +
6
[423(x1)] = x1 +
2223
[429(x1)] = x1 +
1197
[435(x1)] = x1 +
6
[30(x1)] = x1 +
0
[36(x1)] = x1 +
0
[312(x1)] = x1 +
1
[318(x1)] = x1 +
0
[324(x1)] = x1 +
2166
[330(x1)] = x1 +
2223
[31(x1)] = x1 +
2223
[37(x1)] = x1 +
2224
[313(x1)] = x1 +
0
[319(x1)] = x1 +
1824
[325(x1)] = x1 +
6
[331(x1)] = x1 +
685
[32(x1)] = x1 +
0
[38(x1)] = x1 +
0
[314(x1)] = x1 +
0
[320(x1)] = x1 +
2223
[326(x1)] = x1 +
0
[332(x1)] = x1 +
399
[33(x1)] = x1 +
0
[39(x1)] = x1 +
0
[315(x1)] = x1 +
2223
[321(x1)] = x1 +
0
[327(x1)] = x1 +
0
[333(x1)] = x1 +
0
[34(x1)] = x1 +
0
[310(x1)] = x1 +
0
[316(x1)] = x1 +
0
[322(x1)] = x1 +
1
[328(x1)] = x1 +
6
[334(x1)] = x1 +
2223
[35(x1)] = x1 +
2223
[311(x1)] = x1 +
0
[317(x1)] = x1 +
0
[323(x1)] = x1 +
1767
[329(x1)] = x1 +
2223
[335(x1)] = x1 +
2223
[20(x1)] = x1 +
1368
[26(x1)] = x1 +
2223
[212(x1)] = x1 +
2224
[218(x1)] = x1 +
2223
[224(x1)] = x1 +
2229
[230(x1)] = x1 +
2229
[21(x1)] = x1 +
2223
[27(x1)] = x1 +
0
[213(x1)] = x1 +
1881
[219(x1)] = x1 +
6
[225(x1)] = x1 +
2223
[231(x1)] = x1 +
2223
[22(x1)] = x1 +
0
[28(x1)] = x1 +
6
[214(x1)] = x1 +
0
[220(x1)] = x1 +
2223
[226(x1)] = x1 +
2223
[232(x1)] = x1 +
2223
[23(x1)] = x1 +
1
[29(x1)] = x1 +
2223
[215(x1)] = x1 +
2223
[221(x1)] = x1 +
0
[227(x1)] = x1 +
2223
[233(x1)] = x1 +
855
[24(x1)] = x1 +
1
[210(x1)] = x1 +
2223
[216(x1)] = x1 +
0
[222(x1)] = x1 +
2223
[228(x1)] = x1 +
2052
[234(x1)] = x1 +
2229
[25(x1)] = x1 +
2223
[211(x1)] = x1 +
6
[217(x1)] = x1 +
0
[223(x1)] = x1 +
1767
[229(x1)] = x1 +
2223
[235(x1)] = x1 +
2223
[10(x1)] = x1 +
2223
[16(x1)] = x1 +
1
[112(x1)] = x1 +
2223
[118(x1)] = x1 +
1
[124(x1)] = x1 +
2223
[130(x1)] = x1 +
2223
[11(x1)] = x1 +
2229
[17(x1)] = x1 +
0
[113(x1)] = x1 +
0
[119(x1)] = x1 +
0
[125(x1)] = x1 +
2223
[131(x1)] = x1 +
6
[12(x1)] = x1 +
2223
[18(x1)] = x1 +
0
[114(x1)] = x1 +
0
[120(x1)] = x1 +
0
[126(x1)] = x1 +
0
[132(x1)] = x1 +
2223
[13(x1)] = x1 +
2223
[19(x1)] = x1 +
6
[115(x1)] = x1 +
6
[121(x1)] = x1 +
1368
[127(x1)] = x1 +
0
[133(x1)] = x1 +
1368
[14(x1)] = x1 +
2229
[110(x1)] = x1 +
0
[116(x1)] = x1 +
0
[122(x1)] = x1 +
6
[128(x1)] = x1 +
2223
[134(x1)] = x1 +
2229
[15(x1)] = x1 +
2229
[111(x1)] = x1 +
2229
[117(x1)] = x1 +
2223
[123(x1)] = x1 +
2223
[129(x1)] = x1 +
2229
[135(x1)] = x1 +
2223
[00(x1)] = x1 +
2224
[06(x1)] = x1 +
6
[012(x1)] = x1 +
2223
[018(x1)] = x1 +
2223
[024(x1)] = x1 +
2229
[030(x1)] = x1 +
2229
[01(x1)] = x1 +
2229
[07(x1)] = x1 +
1539
[013(x1)] = x1 +
6
[019(x1)] = x1 +
348
[025(x1)] = x1 +
6
[031(x1)] = x1 +
0
[02(x1)] = x1 +
1
[08(x1)] = x1 +
2223
[014(x1)] = x1 +
0
[020(x1)] = x1 +
2223
[026(x1)] = x1 +
0
[032(x1)] = x1 +
2229
[03(x1)] = x1 +
2223
[09(x1)] = x1 +
2223
[015(x1)] = x1 +
0
[021(x1)] = x1 +
1026
[027(x1)] = x1 +
2223
[033(x1)] = x1 +
6
[04(x1)] = x1 +
1
[010(x1)] = x1 +
2223
[016(x1)] = x1 +
2223
[022(x1)] = x1 +
2224
[028(x1)] = x1 +
6
[034(x1)] = x1 +
2223
[05(x1)] = x1 +
2223
[011(x1)] = x1 +
0
[017(x1)] = x1 +
2223
[023(x1)] = x1 +
6
[029(x1)] = x1 +
2223
[035(x1)] = x1 +
2223
all of the following rules can be deleted.

There are 51840 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.