Certification Problem
Input (TPDB SRS_Relative/Waldmann_19/random-70)
The relative rewrite relation R/S is considered where R is the following TRS
b(a(a(x1))) |
→ |
c(b(b(x1))) |
(1) |
a(a(a(x1))) |
→ |
a(c(c(x1))) |
(2) |
a(c(b(x1))) |
→ |
a(a(b(x1))) |
(3) |
c(a(c(x1))) |
→ |
c(c(c(x1))) |
(4) |
and S is the following TRS.
a(a(b(x1))) |
→ |
b(a(b(x1))) |
(5) |
c(a(a(x1))) |
→ |
c(a(a(x1))) |
(6) |
c(c(a(x1))) |
→ |
c(b(b(x1))) |
(7) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 String Reversal
Since only unary symbols occur, one can reverse all terms and obtains the TRS
a(a(b(x1))) |
→ |
b(b(c(x1))) |
(8) |
a(a(a(x1))) |
→ |
c(c(a(x1))) |
(9) |
b(c(a(x1))) |
→ |
b(a(a(x1))) |
(10) |
c(a(c(x1))) |
→ |
c(c(c(x1))) |
(4) |
b(a(a(x1))) |
→ |
b(a(b(x1))) |
(11) |
a(c(c(x1))) |
→ |
b(b(c(x1))) |
(12) |
1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
[a(x1)] |
= |
1 + 1 · x1
|
[b(x1)] |
= |
1 · x1
|
[c(x1)] |
= |
1 + 1 · x1
|
all of the following rules can be deleted.
a(a(b(x1))) |
→ |
b(b(c(x1))) |
(8) |
b(a(a(x1))) |
→ |
b(a(b(x1))) |
(11) |
a(c(c(x1))) |
→ |
b(b(c(x1))) |
(12) |
1.1.1 String Reversal
Since only unary symbols occur, one can reverse all terms and obtains the TRS
a(a(a(x1))) |
→ |
a(c(c(x1))) |
(2) |
a(c(b(x1))) |
→ |
a(a(b(x1))) |
(3) |
c(a(c(x1))) |
→ |
c(c(c(x1))) |
(4) |
1.1.1.1 Bounds
The given TRS is
match-(raise)-bounded by 2.
This is shown by the following automaton.
-
final states:
{0, 1, 2}
-
transitions:
a0(0) |
→ |
0 |
a0(1) |
→ |
0 |
a0(2) |
→ |
0 |
c0(0) |
→ |
1 |
c0(1) |
→ |
1 |
c0(2) |
→ |
1 |
b0(0) |
→ |
2 |
b0(1) |
→ |
2 |
b0(2) |
→ |
2 |
c1(0) |
→ |
4 |
c1(4) |
→ |
3 |
a1(3) |
→ |
0 |
c1(1) |
→ |
4 |
c1(2) |
→ |
4 |
b1(0) |
→ |
5 |
a1(5) |
→ |
3 |
b1(1) |
→ |
5 |
b1(2) |
→ |
5 |
c1(3) |
→ |
1 |
c1(5) |
→ |
0 |
c2(4) |
→ |
7 |
c2(7) |
→ |
6 |
c2(6) |
→ |
4 |
c2(0) |
→ |
7 |
c2(1) |
→ |
7 |
c2(2) |
→ |
7 |
c1(6) |
→ |
0 |
4 |
→ |
3 |
4 |
→ |
0 |
4 |
→ |
7 |
7 |
→ |
6 |