Certification Problem

Input (TPDB SRS_Standard/ICFP_2010/58221)

The rewrite relation of the following TRS is considered.

0(1(2(1(x1)))) → 0(3(4(1(x1)))) (1)
1(3(5(5(x1)))) → 3(1(2(x1))) (2)
2(0(2(3(4(x1))))) → 2(4(4(1(x1)))) (3)
2(3(4(3(2(x1))))) → 0(3(1(3(x1)))) (4)
4(1(0(3(0(0(x1)))))) → 4(0(5(3(2(0(x1)))))) (5)
5(0(2(4(1(5(x1)))))) → 5(4(4(2(4(x1))))) (6)
0(5(3(5(0(2(3(x1))))))) → 4(0(1(1(1(3(x1)))))) (7)
4(3(1(4(2(4(0(x1))))))) → 1(0(3(5(2(1(4(x1))))))) (8)
2(0(3(3(5(5(0(3(x1)))))))) → 2(0(0(3(0(1(3(x1))))))) (9)
3(5(5(0(0(3(5(4(x1)))))))) → 3(4(4(1(2(0(2(x1))))))) (10)
4(3(1(3(4(4(1(5(x1)))))))) → 4(5(2(3(2(4(5(5(x1)))))))) (11)
4(1(1(4(5(0(0(1(3(x1))))))))) → 4(2(3(3(3(2(3(1(3(x1))))))))) (12)
5(2(1(5(3(1(3(1(5(x1))))))))) → 5(5(0(2(1(5(4(5(5(x1))))))))) (13)
0(1(1(3(1(4(4(5(0(0(x1)))))))))) → 5(3(2(3(4(4(4(1(2(4(x1)))))))))) (14)
1(1(5(3(1(2(4(4(0(4(x1)))))))))) → 3(1(2(1(4(2(2(0(4(4(x1)))))))))) (15)
5(5(1(3(2(2(2(2(5(3(x1)))))))))) → 5(2(1(2(2(0(2(3(3(x1))))))))) (16)
0(5(0(3(1(3(1(5(1(5(1(x1))))))))))) → 0(2(5(3(2(0(0(4(1(2(0(x1))))))))))) (17)
1(1(1(3(3(3(4(4(3(1(0(x1))))))))))) → 3(2(5(1(3(1(5(0(0(4(2(x1))))))))))) (18)
1(4(4(3(1(0(1(2(0(1(5(x1))))))))))) → 4(3(3(5(0(5(3(4(5(5(3(5(x1)))))))))))) (19)
2(5(4(0(3(1(5(5(1(5(5(x1))))))))))) → 2(4(0(1(3(4(5(4(4(5(x1)))))))))) (20)
4(1(3(4(1(0(4(0(4(0(3(x1))))))))))) → 4(3(2(4(5(2(4(2(5(3(3(x1))))))))))) (21)
5(0(1(4(5(3(0(5(1(1(0(x1))))))))))) → 5(5(3(0(0(0(5(3(3(3(1(x1))))))))))) (22)
2(4(1(3(3(1(2(4(0(1(2(0(2(x1))))))))))))) → 4(0(3(2(5(1(0(1(1(1(0(0(2(x1))))))))))))) (23)
3(4(1(4(5(2(1(5(0(1(4(2(2(x1))))))))))))) → 5(3(5(5(1(1(1(4(2(0(2(0(0(x1))))))))))))) (24)
4(0(2(2(2(1(2(4(3(1(0(4(3(x1))))))))))))) → 3(3(2(0(1(2(0(3(2(0(4(0(3(x1))))))))))))) (25)
0(2(1(1(5(0(4(0(5(2(3(3(4(4(0(x1))))))))))))))) → 0(5(1(2(3(2(1(3(3(1(1(0(3(5(4(x1))))))))))))))) (26)
3(1(0(5(2(0(2(0(3(1(4(2(3(0(5(x1))))))))))))))) → 3(3(1(2(4(3(3(1(5(5(2(0(5(3(5(x1))))))))))))))) (27)
2(0(5(3(0(4(1(2(5(5(3(2(4(2(5(2(x1)))))))))))))))) → 5(3(4(4(3(3(3(3(1(4(5(1(1(5(0(x1))))))))))))))) (28)
4(1(2(5(5(5(5(4(2(2(4(2(4(4(2(2(x1)))))))))))))))) → 4(5(1(0(1(4(2(1(0(4(2(5(3(2(4(5(x1)))))))))))))))) (29)
0(3(0(5(4(2(2(0(0(5(3(5(0(1(2(2(0(x1))))))))))))))))) → 0(2(5(5(2(3(3(4(0(4(4(5(3(3(5(5(2(x1))))))))))))))))) (30)
4(3(2(2(1(4(2(1(4(0(5(5(5(1(1(4(3(x1))))))))))))))))) → 3(3(1(3(1(2(1(1(0(5(1(3(1(3(4(3(x1)))))))))))))))) (31)
0(3(0(0(5(3(5(5(2(0(4(0(2(1(5(4(3(2(x1)))))))))))))))))) → 2(5(0(2(1(0(0(5(5(2(4(0(5(2(0(3(3(x1))))))))))))))))) (32)
5(1(2(1(5(1(4(0(3(1(2(1(5(1(4(2(3(3(x1)))))))))))))))))) → 5(0(0(2(3(3(2(0(0(5(0(2(0(0(1(2(3(2(x1)))))))))))))))))) (33)
3(0(5(2(2(2(4(3(4(3(4(1(0(1(3(0(4(2(0(x1))))))))))))))))))) → 5(3(4(2(2(4(1(4(2(3(1(5(4(0(4(4(1(2(x1)))))))))))))))))) (34)
5(2(1(5(2(3(4(1(1(1(2(3(3(1(3(3(4(1(3(x1))))))))))))))))))) → 0(4(2(1(0(5(3(5(0(5(0(1(2(4(2(1(2(3(x1)))))))))))))))))) (35)
0(1(0(1(4(5(2(2(5(0(0(1(4(4(2(4(0(4(0(0(x1)))))))))))))))))))) → 1(2(4(0(3(3(4(2(4(4(1(0(4(0(3(2(2(1(1(4(x1)))))))))))))))))))) (36)
2(1(4(3(4(0(1(1(0(3(4(0(2(1(2(0(4(4(1(0(x1)))))))))))))))))))) → 4(1(4(4(5(4(2(2(2(5(5(3(4(4(0(2(5(5(0(1(x1)))))))))))))))))))) (37)
5(3(1(0(2(0(0(5(0(2(1(0(5(5(1(5(3(2(0(1(x1)))))))))))))))))))) → 5(3(3(4(2(3(5(3(4(1(0(0(0(2(3(0(2(4(5(0(x1)))))))))))))))))))) (38)
0(1(4(1(4(5(2(0(1(4(5(5(2(0(3(5(3(4(4(1(4(x1))))))))))))))))))))) → 5(3(0(1(1(2(4(5(1(0(0(5(5(3(4(3(1(3(2(0(4(x1))))))))))))))))))))) (39)
1(5(0(4(1(1(3(5(4(3(4(4(3(1(5(4(0(1(3(3(3(x1))))))))))))))))))))) → 3(5(1(0(5(1(0(3(1(3(2(2(1(0(4(4(2(1(3(0(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 +
24245/19
[56(x1)] = x1 +
13130/19
[512(x1)] = x1 +
0
[518(x1)] = x1 +
2034/19
[524(x1)] = x1 +
24245/19
[530(x1)] = x1 +
13130/19
[51(x1)] = x1 +
0
[57(x1)] = x1 +
0
[513(x1)] = x1 +
24245/19
[519(x1)] = x1 +
6
[525(x1)] = x1 +
6
[531(x1)] = x1 +
24245/19
[52(x1)] = x1 +
24264/19
[58(x1)] = x1 +
6
[514(x1)] = x1 +
0
[520(x1)] = x1 +
0
[526(x1)] = x1 +
24264/19
[532(x1)] = x1 +
24245/19
[53(x1)] = x1 +
0
[59(x1)] = x1 +
6
[515(x1)] = x1 +
20865/19
[521(x1)] = x1 +
24245/19
[527(x1)] = x1 +
24245/19
[533(x1)] = x1 +
24245/19
[54(x1)] = x1 +
24359/19
[510(x1)] = x1 +
21905/19
[516(x1)] = x1 +
65
[522(x1)] = x1 +
6760/19
[528(x1)] = x1 +
6
[534(x1)] = x1 +
1
[55(x1)] = x1 +
6
[511(x1)] = x1 +
24359/19
[517(x1)] = x1 +
22880/19
[523(x1)] = x1 +
0
[529(x1)] = x1 +
24359/19
[535(x1)] = x1 +
24245/19
[40(x1)] = x1 +
0
[46(x1)] = x1 +
0
[412(x1)] = x1 +
6
[418(x1)] = x1 +
24245/19
[424(x1)] = x1 +
1
[430(x1)] = x1 +
24245/19
[41(x1)] = x1 +
0
[47(x1)] = x1 +
0
[413(x1)] = x1 +
24264/19
[419(x1)] = x1 +
0
[425(x1)] = x1 +
0
[431(x1)] = x1 +
2340/19
[42(x1)] = x1 +
6
[48(x1)] = x1 +
24245/19
[414(x1)] = x1 +
0
[420(x1)] = x1 +
24245/19
[426(x1)] = x1 +
24264/19
[432(x1)] = x1 +
24359/19
[43(x1)] = x1 +
17550/19
[49(x1)] = x1 +
24245/19
[415(x1)] = x1 +
6
[421(x1)] = x1 +
0
[427(x1)] = x1 +
0
[433(x1)] = x1 +
24245/19
[44(x1)] = x1 +
24245/19
[410(x1)] = x1 +
6
[416(x1)] = x1 +
24245/19
[422(x1)] = x1 +
1
[428(x1)] = x1 +
24245/19
[434(x1)] = x1 +
24245/19
[45(x1)] = x1 +
1
[411(x1)] = x1 +
24359/19
[417(x1)] = x1 +
1
[423(x1)] = x1 +
24359/19
[429(x1)] = x1 +
24245/19
[435(x1)] = x1 +
24245/19
[30(x1)] = x1 +
24245/19
[36(x1)] = x1 +
24245/19
[312(x1)] = x1 +
0
[318(x1)] = x1 +
2015/19
[324(x1)] = x1 +
6
[330(x1)] = x1 +
24245/19
[31(x1)] = x1 +
24245/19
[37(x1)] = x1 +
6584/19
[313(x1)] = x1 +
24264/19
[319(x1)] = x1 +
6
[325(x1)] = x1 +
24245/19
[331(x1)] = x1 +
24245/19
[32(x1)] = x1 +
6
[38(x1)] = x1 +
1
[314(x1)] = x1 +
1
[320(x1)] = x1 +
0
[326(x1)] = x1 +
0
[332(x1)] = x1 +
24245/19
[33(x1)] = x1 +
0
[39(x1)] = x1 +
24245/19
[315(x1)] = x1 +
6
[321(x1)] = x1 +
6
[327(x1)] = x1 +
24245/19
[333(x1)] = x1 +
1
[34(x1)] = x1 +
0
[310(x1)] = x1 +
24264/19
[316(x1)] = x1 +
0
[322(x1)] = x1 +
1
[328(x1)] = x1 +
0
[334(x1)] = x1 +
24245/19
[35(x1)] = x1 +
24245/19
[311(x1)] = x1 +
24245/19
[317(x1)] = x1 +
0
[323(x1)] = x1 +
6
[329(x1)] = x1 +
1384/19
[335(x1)] = x1 +
24245/19
[20(x1)] = x1 +
6
[26(x1)] = x1 +
24245/19
[212(x1)] = x1 +
0
[218(x1)] = x1 +
24245/19
[224(x1)] = x1 +
0
[230(x1)] = x1 +
24245/19
[21(x1)] = x1 +
0
[27(x1)] = x1 +
0
[213(x1)] = x1 +
24245/19
[219(x1)] = x1 +
24245/19
[225(x1)] = x1 +
6
[231(x1)] = x1 +
11979/19
[22(x1)] = x1 +
24359/19
[28(x1)] = x1 +
24264/19
[214(x1)] = x1 +
18934/19
[220(x1)] = x1 +
0
[226(x1)] = x1 +
1
[232(x1)] = x1 +
24264/19
[23(x1)] = x1 +
24359/19
[29(x1)] = x1 +
24245/19
[215(x1)] = x1 +
24245/19
[221(x1)] = x1 +
24245/19
[227(x1)] = x1 +
6
[233(x1)] = x1 +
1
[24(x1)] = x1 +
24359/19
[210(x1)] = x1 +
5525/19
[216(x1)] = x1 +
1
[222(x1)] = x1 +
6565/19
[228(x1)] = x1 +
1
[234(x1)] = x1 +
1
[25(x1)] = x1 +
21775/19
[211(x1)] = x1 +
24245/19
[217(x1)] = x1 +
22880/19
[223(x1)] = x1 +
1
[229(x1)] = x1 +
24264/19
[235(x1)] = x1 +
1
[10(x1)] = x1 +
24245/19
[16(x1)] = x1 +
13130/19
[112(x1)] = x1 +
24264/19
[118(x1)] = x1 +
24359/19
[124(x1)] = x1 +
24245/19
[130(x1)] = x1 +
24245/19
[11(x1)] = x1 +
24245/19
[17(x1)] = x1 +
24245/19
[113(x1)] = x1 +
24245/19
[119(x1)] = x1 +
19844/19
[125(x1)] = x1 +
24359/19
[131(x1)] = x1 +
2340/19
[12(x1)] = x1 +
24245/19
[18(x1)] = x1 +
14319/19
[114(x1)] = x1 +
18915/19
[120(x1)] = x1 +
0
[126(x1)] = x1 +
0
[132(x1)] = x1 +
24245/19
[13(x1)] = x1 +
24359/19
[19(x1)] = x1 +
0
[115(x1)] = x1 +
1
[121(x1)] = x1 +
0
[127(x1)] = x1 +
24264/19
[133(x1)] = x1 +
15600/19
[14(x1)] = x1 +
24359/19
[110(x1)] = x1 +
6
[116(x1)] = x1 +
24245/19
[122(x1)] = x1 +
6
[128(x1)] = x1 +
0
[134(x1)] = x1 +
24245/19
[15(x1)] = x1 +
1
[111(x1)] = x1 +
1
[117(x1)] = x1 +
0
[123(x1)] = x1 +
24245/19
[129(x1)] = x1 +
24245/19
[135(x1)] = x1 +
1
[00(x1)] = x1 +
6
[06(x1)] = x1 +
24245/19
[012(x1)] = x1 +
24264/19
[018(x1)] = x1 +
24245/19
[024(x1)] = x1 +
23010/19
[030(x1)] = x1 +
24245/19
[01(x1)] = x1 +
6
[07(x1)] = x1 +
1
[013(x1)] = x1 +
24359/19
[019(x1)] = x1 +
0
[025(x1)] = x1 +
24245/19
[031(x1)] = x1 +
24245/19
[02(x1)] = x1 +
0
[08(x1)] = x1 +
1365/19
[014(x1)] = x1 +
0
[020(x1)] = x1 +
1
[026(x1)] = x1 +
0
[032(x1)] = x1 +
24245/19
[03(x1)] = x1 +
6
[09(x1)] = x1 +
24245/19
[015(x1)] = x1 +
24264/19
[021(x1)] = x1 +
24245/19
[027(x1)] = x1 +
24245/19
[033(x1)] = x1 +
1
[04(x1)] = x1 +
24245/19
[010(x1)] = x1 +
24245/19
[016(x1)] = x1 +
24245/19
[022(x1)] = x1 +
24264/19
[028(x1)] = x1 +
24245/19
[034(x1)] = x1 +
24245/19
[05(x1)] = x1 +
24264/19
[011(x1)] = x1 +
0
[017(x1)] = x1 +
9035/19
[023(x1)] = x1 +
0
[029(x1)] = x1 +
24245/19
[035(x1)] = x1 +
11895/19
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.