Certification Problem
Input (TPDB SRS_Relative/Zantema_06_relative/rel10)
The relative rewrite relation R/S is considered where R is the following TRS
b(p(b(x1))) |
→ |
b(q(b(x1))) |
(1) |
and S is the following TRS.
q(x1) |
→ |
1(q(1(x1))) |
(2) |
1(p(0(1(0(x1))))) |
→ |
p(x1) |
(3) |
q(x1) |
→ |
1(p(1(x1))) |
(4) |
q(x1) |
→ |
0(q(0(x1))) |
(5) |
q(x1) |
→ |
0(p(0(x1))) |
(6) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Closure Under Flat Contexts
Using the flat contexts
{b(☐), p(☐), q(☐), 1(☐), 0(☐)}
We obtain the transformed TRS
b(p(b(x1))) |
→ |
b(q(b(x1))) |
(1) |
b(q(x1)) |
→ |
b(1(q(1(x1)))) |
(7) |
p(q(x1)) |
→ |
p(1(q(1(x1)))) |
(8) |
q(q(x1)) |
→ |
q(1(q(1(x1)))) |
(9) |
1(q(x1)) |
→ |
1(1(q(1(x1)))) |
(10) |
0(q(x1)) |
→ |
0(1(q(1(x1)))) |
(11) |
b(1(p(0(1(0(x1)))))) |
→ |
b(p(x1)) |
(12) |
p(1(p(0(1(0(x1)))))) |
→ |
p(p(x1)) |
(13) |
q(1(p(0(1(0(x1)))))) |
→ |
q(p(x1)) |
(14) |
1(1(p(0(1(0(x1)))))) |
→ |
1(p(x1)) |
(15) |
0(1(p(0(1(0(x1)))))) |
→ |
0(p(x1)) |
(16) |
b(q(x1)) |
→ |
b(1(p(1(x1)))) |
(17) |
p(q(x1)) |
→ |
p(1(p(1(x1)))) |
(18) |
q(q(x1)) |
→ |
q(1(p(1(x1)))) |
(19) |
1(q(x1)) |
→ |
1(1(p(1(x1)))) |
(20) |
0(q(x1)) |
→ |
0(1(p(1(x1)))) |
(21) |
b(q(x1)) |
→ |
b(0(q(0(x1)))) |
(22) |
p(q(x1)) |
→ |
p(0(q(0(x1)))) |
(23) |
q(q(x1)) |
→ |
q(0(q(0(x1)))) |
(24) |
1(q(x1)) |
→ |
1(0(q(0(x1)))) |
(25) |
0(q(x1)) |
→ |
0(0(q(0(x1)))) |
(26) |
b(q(x1)) |
→ |
b(0(p(0(x1)))) |
(27) |
p(q(x1)) |
→ |
p(0(p(0(x1)))) |
(28) |
q(q(x1)) |
→ |
q(0(p(0(x1)))) |
(29) |
1(q(x1)) |
→ |
1(0(p(0(x1)))) |
(30) |
0(q(x1)) |
→ |
0(0(p(0(x1)))) |
(31) |
1.1 Semantic Labeling
Root-labeling is applied.
We obtain the labeled TRS
There are 130 ruless (increase limit for explicit display).
1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
[bp(x1)] |
= |
1 · x1
|
[pb(x1)] |
= |
1 + 1 · x1
|
[bb(x1)] |
= |
1 · x1
|
[bq(x1)] |
= |
1 · x1
|
[qb(x1)] |
= |
1 + 1 · x1
|
[b1(x1)] |
= |
1 · x1
|
[b0(x1)] |
= |
1 · x1
|
[1q(x1)] |
= |
1 · x1
|
[q1(x1)] |
= |
1 + 1 · x1
|
[1b(x1)] |
= |
1 · x1
|
[qp(x1)] |
= |
1 + 1 · x1
|
[1p(x1)] |
= |
1 · x1
|
[qq(x1)] |
= |
1 + 1 · x1
|
[11(x1)] |
= |
1 · x1
|
[q0(x1)] |
= |
1 + 1 · x1
|
[10(x1)] |
= |
1 · x1
|
[pq(x1)] |
= |
1 + 1 · x1
|
[p1(x1)] |
= |
1 · x1
|
[0q(x1)] |
= |
1 · x1
|
[01(x1)] |
= |
1 · x1
|
[p0(x1)] |
= |
1 + 1 · x1
|
[0b(x1)] |
= |
1 · x1
|
[0p(x1)] |
= |
1 · x1
|
[pp(x1)] |
= |
1 · x1
|
[00(x1)] |
= |
1 · x1
|
all of the following rules can be deleted.
pq(qb(x1)) |
→ |
p1(1q(q1(1b(x1)))) |
(42) |
pq(qp(x1)) |
→ |
p1(1q(q1(1p(x1)))) |
(43) |
pq(qq(x1)) |
→ |
p1(1q(q1(1q(x1)))) |
(44) |
pq(q1(x1)) |
→ |
p1(1q(q1(11(x1)))) |
(45) |
pq(q0(x1)) |
→ |
p1(1q(q1(10(x1)))) |
(46) |
b1(1p(p0(01(10(0p(x1)))))) |
→ |
bp(pp(x1)) |
(63) |
b1(1p(p0(01(10(01(x1)))))) |
→ |
bp(p1(x1)) |
(65) |
p1(1p(p0(01(10(0p(x1)))))) |
→ |
pp(pp(x1)) |
(68) |
p1(1p(p0(01(10(01(x1)))))) |
→ |
pp(p1(x1)) |
(70) |
q1(1p(p0(01(10(0p(x1)))))) |
→ |
qp(pp(x1)) |
(73) |
q1(1p(p0(01(10(01(x1)))))) |
→ |
qp(p1(x1)) |
(75) |
11(1p(p0(01(10(0p(x1)))))) |
→ |
1p(pp(x1)) |
(78) |
11(1p(p0(01(10(01(x1)))))) |
→ |
1p(p1(x1)) |
(80) |
01(1p(p0(01(10(0p(x1)))))) |
→ |
0p(pp(x1)) |
(83) |
01(1p(p0(01(10(01(x1)))))) |
→ |
0p(p1(x1)) |
(85) |
bq(qb(x1)) |
→ |
b1(1p(p1(1b(x1)))) |
(87) |
bq(qp(x1)) |
→ |
b1(1p(p1(1p(x1)))) |
(88) |
bq(qq(x1)) |
→ |
b1(1p(p1(1q(x1)))) |
(89) |
bq(q1(x1)) |
→ |
b1(1p(p1(11(x1)))) |
(90) |
bq(q0(x1)) |
→ |
b1(1p(p1(10(x1)))) |
(91) |
pq(qb(x1)) |
→ |
p1(1p(p1(1b(x1)))) |
(92) |
pq(qp(x1)) |
→ |
p1(1p(p1(1p(x1)))) |
(93) |
pq(qq(x1)) |
→ |
p1(1p(p1(1q(x1)))) |
(94) |
pq(q1(x1)) |
→ |
p1(1p(p1(11(x1)))) |
(95) |
pq(q0(x1)) |
→ |
p1(1p(p1(10(x1)))) |
(96) |
qq(qb(x1)) |
→ |
q1(1p(p1(1b(x1)))) |
(97) |
qq(qp(x1)) |
→ |
q1(1p(p1(1p(x1)))) |
(98) |
qq(qq(x1)) |
→ |
q1(1p(p1(1q(x1)))) |
(99) |
qq(q1(x1)) |
→ |
q1(1p(p1(11(x1)))) |
(100) |
qq(q0(x1)) |
→ |
q1(1p(p1(10(x1)))) |
(101) |
1q(qb(x1)) |
→ |
11(1p(p1(1b(x1)))) |
(102) |
1q(qp(x1)) |
→ |
11(1p(p1(1p(x1)))) |
(103) |
1q(qq(x1)) |
→ |
11(1p(p1(1q(x1)))) |
(104) |
1q(q1(x1)) |
→ |
11(1p(p1(11(x1)))) |
(105) |
1q(q0(x1)) |
→ |
11(1p(p1(10(x1)))) |
(106) |
0q(qb(x1)) |
→ |
01(1p(p1(1b(x1)))) |
(107) |
0q(qp(x1)) |
→ |
01(1p(p1(1p(x1)))) |
(108) |
0q(qq(x1)) |
→ |
01(1p(p1(1q(x1)))) |
(109) |
0q(q1(x1)) |
→ |
01(1p(p1(11(x1)))) |
(110) |
0q(q0(x1)) |
→ |
01(1p(p1(10(x1)))) |
(111) |
1.1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
[bp(x1)] |
= |
1 + 1 · x1
|
[pb(x1)] |
= |
1 · x1
|
[bb(x1)] |
= |
1 · x1
|
[bq(x1)] |
= |
1 · x1
|
[qb(x1)] |
= |
1 + 1 · x1
|
[b1(x1)] |
= |
1 · x1
|
[b0(x1)] |
= |
1 · x1
|
[1q(x1)] |
= |
1 · x1
|
[q1(x1)] |
= |
1 · x1
|
[1b(x1)] |
= |
1 · x1
|
[qp(x1)] |
= |
1 + 1 · x1
|
[1p(x1)] |
= |
1 + 1 · x1
|
[qq(x1)] |
= |
1 + 1 · x1
|
[11(x1)] |
= |
1 · x1
|
[q0(x1)] |
= |
1 · x1
|
[10(x1)] |
= |
1 · x1
|
[0q(x1)] |
= |
1 · x1
|
[01(x1)] |
= |
1 · x1
|
[p0(x1)] |
= |
1 · x1
|
[0b(x1)] |
= |
1 · x1
|
[pq(x1)] |
= |
1 · x1
|
[00(x1)] |
= |
1 · x1
|
[p1(x1)] |
= |
1 + 1 · x1
|
[pp(x1)] |
= |
1 + 1 · x1
|
[0p(x1)] |
= |
1 · x1
|
all of the following rules can be deleted.
bq(qb(x1)) |
→ |
b1(1q(q1(1b(x1)))) |
(37) |
bq(qq(x1)) |
→ |
b1(1q(q1(1q(x1)))) |
(39) |
qq(qb(x1)) |
→ |
q1(1q(q1(1b(x1)))) |
(47) |
qq(qp(x1)) |
→ |
q1(1q(q1(1p(x1)))) |
(48) |
qq(qq(x1)) |
→ |
q1(1q(q1(1q(x1)))) |
(49) |
qq(q1(x1)) |
→ |
q1(1q(q1(11(x1)))) |
(50) |
qq(q0(x1)) |
→ |
q1(1q(q1(10(x1)))) |
(51) |
1q(qb(x1)) |
→ |
11(1q(q1(1b(x1)))) |
(52) |
1q(qq(x1)) |
→ |
11(1q(q1(1q(x1)))) |
(54) |
0q(qb(x1)) |
→ |
01(1q(q1(1b(x1)))) |
(57) |
0q(qq(x1)) |
→ |
01(1q(q1(1q(x1)))) |
(59) |
p1(1p(p0(01(10(0b(x1)))))) |
→ |
pp(pb(x1)) |
(67) |
p1(1p(p0(01(10(0q(x1)))))) |
→ |
pp(pq(x1)) |
(69) |
p1(1p(p0(01(10(00(x1)))))) |
→ |
pp(p0(x1)) |
(71) |
01(1p(p0(01(10(0b(x1)))))) |
→ |
0p(pb(x1)) |
(82) |
01(1p(p0(01(10(0q(x1)))))) |
→ |
0p(pq(x1)) |
(84) |
01(1p(p0(01(10(00(x1)))))) |
→ |
0p(p0(x1)) |
(86) |
bq(qb(x1)) |
→ |
b0(0q(q0(0b(x1)))) |
(112) |
bq(qp(x1)) |
→ |
b0(0q(q0(0p(x1)))) |
(113) |
bq(qq(x1)) |
→ |
b0(0q(q0(0q(x1)))) |
(114) |
pq(qb(x1)) |
→ |
p0(0q(q0(0b(x1)))) |
(117) |
pq(qp(x1)) |
→ |
p0(0q(q0(0p(x1)))) |
(118) |
pq(qq(x1)) |
→ |
p0(0q(q0(0q(x1)))) |
(119) |
qq(qb(x1)) |
→ |
q0(0q(q0(0b(x1)))) |
(122) |
qq(qp(x1)) |
→ |
q0(0q(q0(0p(x1)))) |
(123) |
qq(qq(x1)) |
→ |
q0(0q(q0(0q(x1)))) |
(124) |
qq(q1(x1)) |
→ |
q0(0q(q0(01(x1)))) |
(125) |
qq(q0(x1)) |
→ |
q0(0q(q0(00(x1)))) |
(126) |
1q(qb(x1)) |
→ |
10(0q(q0(0b(x1)))) |
(127) |
1q(qp(x1)) |
→ |
10(0q(q0(0p(x1)))) |
(128) |
1q(qq(x1)) |
→ |
10(0q(q0(0q(x1)))) |
(129) |
0q(qb(x1)) |
→ |
00(0q(q0(0b(x1)))) |
(132) |
0q(qp(x1)) |
→ |
00(0q(q0(0p(x1)))) |
(133) |
0q(qq(x1)) |
→ |
00(0q(q0(0q(x1)))) |
(134) |
bq(qb(x1)) |
→ |
b0(0p(p0(0b(x1)))) |
(137) |
bq(qp(x1)) |
→ |
b0(0p(p0(0p(x1)))) |
(138) |
bq(qq(x1)) |
→ |
b0(0p(p0(0q(x1)))) |
(139) |
pq(qb(x1)) |
→ |
p0(0p(p0(0b(x1)))) |
(142) |
pq(qp(x1)) |
→ |
p0(0p(p0(0p(x1)))) |
(143) |
pq(qq(x1)) |
→ |
p0(0p(p0(0q(x1)))) |
(144) |
qq(qb(x1)) |
→ |
q0(0p(p0(0b(x1)))) |
(147) |
qq(qp(x1)) |
→ |
q0(0p(p0(0p(x1)))) |
(148) |
qq(qq(x1)) |
→ |
q0(0p(p0(0q(x1)))) |
(149) |
qq(q1(x1)) |
→ |
q0(0p(p0(01(x1)))) |
(150) |
qq(q0(x1)) |
→ |
q0(0p(p0(00(x1)))) |
(151) |
1q(qb(x1)) |
→ |
10(0p(p0(0b(x1)))) |
(152) |
1q(qp(x1)) |
→ |
10(0p(p0(0p(x1)))) |
(153) |
1q(qq(x1)) |
→ |
10(0p(p0(0q(x1)))) |
(154) |
0q(qb(x1)) |
→ |
00(0p(p0(0b(x1)))) |
(157) |
0q(qp(x1)) |
→ |
00(0p(p0(0p(x1)))) |
(158) |
0q(qq(x1)) |
→ |
00(0p(p0(0q(x1)))) |
(159) |
1.1.1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
[bp(x1)] |
= |
1 · x1
|
[pb(x1)] |
= |
1 + 1 · x1
|
[bb(x1)] |
= |
1 · x1
|
[bq(x1)] |
= |
1 + 1 · x1
|
[qb(x1)] |
= |
1 · x1
|
[b1(x1)] |
= |
1 + 1 · x1
|
[b0(x1)] |
= |
1 · x1
|
[qp(x1)] |
= |
1 · x1
|
[1q(x1)] |
= |
1 · x1
|
[q1(x1)] |
= |
1 · x1
|
[1p(x1)] |
= |
1 · x1
|
[11(x1)] |
= |
1 · x1
|
[q0(x1)] |
= |
1 · x1
|
[10(x1)] |
= |
1 · x1
|
[0q(x1)] |
= |
1 · x1
|
[01(x1)] |
= |
1 · x1
|
[p0(x1)] |
= |
1 · x1
|
[0b(x1)] |
= |
1 + 1 · x1
|
[pq(x1)] |
= |
1 · x1
|
[00(x1)] |
= |
1 · x1
|
[0p(x1)] |
= |
1 · x1
|
all of the following rules can be deleted.
b1(1p(p0(01(10(0b(x1)))))) |
→ |
bp(pb(x1)) |
(62) |
b1(1p(p0(01(10(0q(x1)))))) |
→ |
bp(pq(x1)) |
(64) |
b1(1p(p0(01(10(00(x1)))))) |
→ |
bp(p0(x1)) |
(66) |
bq(q1(x1)) |
→ |
b0(0q(q0(01(x1)))) |
(115) |
bq(q0(x1)) |
→ |
b0(0q(q0(00(x1)))) |
(116) |
bq(q1(x1)) |
→ |
b0(0p(p0(01(x1)))) |
(140) |
bq(q0(x1)) |
→ |
b0(0p(p0(00(x1)))) |
(141) |
1.1.1.1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
[bp(x1)] |
= |
1 + 1 · x1
|
[pb(x1)] |
= |
1 + 1 · x1
|
[bb(x1)] |
= |
1 · x1
|
[bq(x1)] |
= |
1 + 1 · x1
|
[qb(x1)] |
= |
1 · x1
|
[b1(x1)] |
= |
1 · x1
|
[b0(x1)] |
= |
1 · x1
|
[qp(x1)] |
= |
1 · x1
|
[1q(x1)] |
= |
1 · x1
|
[q1(x1)] |
= |
1 · x1
|
[1p(x1)] |
= |
1 · x1
|
[11(x1)] |
= |
1 · x1
|
[q0(x1)] |
= |
1 · x1
|
[10(x1)] |
= |
1 · x1
|
[0q(x1)] |
= |
1 · x1
|
[01(x1)] |
= |
1 · x1
|
[p0(x1)] |
= |
1 · x1
|
[0b(x1)] |
= |
1 + 1 · x1
|
[pq(x1)] |
= |
1 · x1
|
[00(x1)] |
= |
1 · x1
|
[0p(x1)] |
= |
1 · x1
|
all of the following rules can be deleted.
bp(pb(bb(x1))) |
→ |
bq(qb(bb(x1))) |
(32) |
bp(pb(bp(x1))) |
→ |
bq(qb(bp(x1))) |
(33) |
bp(pb(bq(x1))) |
→ |
bq(qb(bq(x1))) |
(34) |
bp(pb(b1(x1))) |
→ |
bq(qb(b1(x1))) |
(35) |
bp(pb(b0(x1))) |
→ |
bq(qb(b0(x1))) |
(36) |
bq(qp(x1)) |
→ |
b1(1q(q1(1p(x1)))) |
(38) |
bq(q1(x1)) |
→ |
b1(1q(q1(11(x1)))) |
(40) |
bq(q0(x1)) |
→ |
b1(1q(q1(10(x1)))) |
(41) |
1.1.1.1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.