Certification Problem
Input (TPDB TRS_Relative/Relative_05/rtL-me3)
The relative rewrite relation R/S is considered where R is the following TRS
|
topB(i,N1(x),y) |
→ |
topA(1,T1(x),y) |
(1) |
|
topA(i,x,N2(y)) |
→ |
topB(0,x,T2(y)) |
(2) |
|
topB(i,S1(x),y) |
→ |
topA(i,N1(x),y) |
(3) |
|
topA(i,x,S2(y)) |
→ |
topB(i,x,N2(y)) |
(4) |
|
topA(i,N1(x),T2(y)) |
→ |
topB(i,N1(x),S2(y)) |
(5) |
|
topA(1,T1(x),T2(y)) |
→ |
topB(1,T1(x),S2(y)) |
(6) |
and S is the following TRS.
|
topB(i,x,N2(y)) |
→ |
topB(i,x,N2(C(y))) |
(7) |
|
topA(i,S1(x),y) |
→ |
topA(i,N1(x),y) |
(8) |
|
topB(i,x,S2(y)) |
→ |
topB(i,x,N2(y)) |
(9) |
|
topB(i,x,N2(y)) |
→ |
topB(0,x,T2(y)) |
(10) |
|
topA(i,T1(x),y) |
→ |
topA(i,T1(x),y) |
(11) |
|
topB(1,T1(x),T2(y)) |
→ |
topB(1,T1(x),S2(y)) |
(12) |
|
topB(i,N1(x),T2(y)) |
→ |
topB(i,N1(x),S2(y)) |
(13) |
|
topA(i,N1(x),y) |
→ |
topA(i,N1(C(x)),y) |
(14) |
|
topB(i,x,T2(y)) |
→ |
topB(i,x,T2(y)) |
(15) |
|
topB(i,x,S2(y)) |
→ |
topB(i,x,S2(D(y))) |
(16) |
|
topA(i,N1(x),y) |
→ |
topA(1,T1(x),y) |
(17) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [topB(x1, x2, x3)] |
= |
1 · x1 + 1 · x2 + 1 · x3
|
| [N1(x1)] |
= |
1 · x1
|
| [topA(x1, x2, x3)] |
= |
1 · x1 + 1 · x2 + 1 · x3
|
| [1] |
= |
0 |
| [T1(x1)] |
= |
1 · x1
|
| [N2(x1)] |
= |
1 · x1
|
| [0] |
= |
0 |
| [T2(x1)] |
= |
1 · x1
|
| [S1(x1)] |
= |
1 + 1 · x1
|
| [S2(x1)] |
= |
1 · x1
|
| [C(x1)] |
= |
1 · x1
|
| [D(x1)] |
= |
1 · x1
|
all of the following rules can be deleted.
|
topB(i,S1(x),y) |
→ |
topA(i,N1(x),y) |
(3) |
|
topA(i,S1(x),y) |
→ |
topA(i,N1(x),y) |
(8) |
|
topA(i,T1(x),y) |
→ |
topA(i,T1(x),y) |
(11) |
|
topB(i,x,T2(y)) |
→ |
topB(i,x,T2(y)) |
(15) |
1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [topB(x1, x2, x3)] |
= |
1 · x1 + 1 · x2 + 1 · x3
|
| [N1(x1)] |
= |
1 + 1 · x1
|
| [topA(x1, x2, x3)] |
= |
1 + 1 · x1 + 1 · x2 + 1 · x3
|
| [1] |
= |
0 |
| [T1(x1)] |
= |
1 · x1
|
| [N2(x1)] |
= |
1 · x1
|
| [0] |
= |
0 |
| [T2(x1)] |
= |
1 · x1
|
| [S2(x1)] |
= |
1 · x1
|
| [C(x1)] |
= |
1 · x1
|
| [D(x1)] |
= |
1 · x1
|
all of the following rules can be deleted.
|
topA(i,x,N2(y)) |
→ |
topB(0,x,T2(y)) |
(2) |
|
topA(i,x,S2(y)) |
→ |
topB(i,x,N2(y)) |
(4) |
|
topA(i,N1(x),T2(y)) |
→ |
topB(i,N1(x),S2(y)) |
(5) |
|
topA(1,T1(x),T2(y)) |
→ |
topB(1,T1(x),S2(y)) |
(6) |
|
topA(i,N1(x),y) |
→ |
topA(1,T1(x),y) |
(17) |
1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [topB(x1, x2, x3)] |
= |
1 · x1 + 1 · x2 + 1 · x3
|
| [N1(x1)] |
= |
1 + 1 · x1
|
| [topA(x1, x2, x3)] |
= |
1 · x1 + 1 · x2 + 1 · x3
|
| [1] |
= |
0 |
| [T1(x1)] |
= |
1 · x1
|
| [N2(x1)] |
= |
1 + 1 · x1
|
| [C(x1)] |
= |
1 · x1
|
| [S2(x1)] |
= |
1 + 1 · x1
|
| [0] |
= |
0 |
| [T2(x1)] |
= |
1 + 1 · x1
|
| [D(x1)] |
= |
1 · x1
|
all of the following rules can be deleted.
|
topB(i,N1(x),y) |
→ |
topA(1,T1(x),y) |
(1) |
1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.