Certification Problem
Input (TPDB TRS_Standard/Transformed_CSR_04/Ex5_Zan97_FR)
The rewrite relation of the following TRS is considered.
|
f(X) |
→ |
if(X,c,n__f(n__true)) |
(1) |
|
if(true,X,Y) |
→ |
X |
(2) |
|
if(false,X,Y) |
→ |
activate(Y) |
(3) |
|
f(X) |
→ |
n__f(X) |
(4) |
| true |
→ |
n__true |
(5) |
|
activate(n__f(X)) |
→ |
f(activate(X)) |
(6) |
|
activate(n__true) |
→ |
true |
(7) |
|
activate(X) |
→ |
X |
(8) |
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
| [f(x1)] |
= |
2 · x1
|
| [if(x1, x2, x3)] |
= |
1 · x1 + 1 · x2 + 2 · x3
|
| [c] |
= |
0 |
| [n__f(x1)] |
= |
2 · x1
|
| [n__true] |
= |
0 |
| [true] |
= |
0 |
| [false] |
= |
1 |
| [activate(x1)] |
= |
1 · x1
|
all of the following rules can be deleted.
|
if(false,X,Y) |
→ |
activate(Y) |
(3) |
1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [f(x1)] |
= |
2 + 2 · x1
|
| [if(x1, x2, x3)] |
= |
1 · x1 + 1 · x2 + 1 · x3
|
| [c] |
= |
0 |
| [n__f(x1)] |
= |
2 + 2 · x1
|
| [n__true] |
= |
0 |
| [true] |
= |
1 |
| [activate(x1)] |
= |
2 + 2 · x1
|
all of the following rules can be deleted.
|
if(true,X,Y) |
→ |
X |
(2) |
| true |
→ |
n__true |
(5) |
|
activate(n__true) |
→ |
true |
(7) |
|
activate(X) |
→ |
X |
(8) |
1.1.1 Rule Removal
Using the
| prec(f) |
= |
2 |
|
stat(f) |
= |
lex
|
| prec(if) |
= |
0 |
|
stat(if) |
= |
lex
|
| prec(c) |
= |
0 |
|
stat(c) |
= |
lex
|
| prec(n__f) |
= |
1 |
|
stat(n__f) |
= |
lex
|
| prec(n__true) |
= |
2 |
|
stat(n__true) |
= |
lex
|
| prec(activate) |
= |
3 |
|
stat(activate) |
= |
lex
|
| π(f) |
= |
[1] |
| π(if) |
= |
[2,3,1] |
| π(c) |
= |
[] |
| π(n__f) |
= |
[1] |
| π(n__true) |
= |
[] |
| π(activate) |
= |
[1] |
all of the following rules can be deleted.
|
f(X) |
→ |
if(X,c,n__f(n__true)) |
(1) |
|
f(X) |
→ |
n__f(X) |
(4) |
|
activate(n__f(X)) |
→ |
f(activate(X)) |
(6) |
1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.