Certification Problem
Input (TPDB TRS_Innermost/Transformed_CSR_innermost_04/Ex9_Luc06_GM)
The rewrite relation of the following TRS is considered.
a__f(a,X,X) |
→ |
a__f(X,a__b,b) |
(1) |
a__b |
→ |
a |
(2) |
mark(f(X1,X2,X3)) |
→ |
a__f(X1,mark(X2),X3) |
(3) |
mark(b) |
→ |
a__b |
(4) |
mark(a) |
→ |
a |
(5) |
a__f(X1,X2,X3) |
→ |
f(X1,X2,X3) |
(6) |
a__b |
→ |
b |
(7) |
The evaluation strategy is innermost.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
[a__f(x1, x2, x3)] |
= |
2 + 2 · x1 + 2 · x2 + 2 · x3
|
[a] |
= |
1 |
[a__b] |
= |
1 |
[b] |
= |
0 |
[mark(x1)] |
= |
1 + 2 · x1
|
[f(x1, x2, x3)] |
= |
2 + 2 · x1 + 2 · x2 + 2 · x3
|
all of the following rules can be deleted.
mark(f(X1,X2,X3)) |
→ |
a__f(X1,mark(X2),X3) |
(3) |
mark(a) |
→ |
a |
(5) |
a__b |
→ |
b |
(7) |
1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
[a__f(x1, x2, x3)] |
= |
1 + 1 · x1 + 2 · x2 + 2 · x3
|
[a] |
= |
0 |
[a__b] |
= |
0 |
[b] |
= |
0 |
[mark(x1)] |
= |
2 + 2 · x1
|
[f(x1, x2, x3)] |
= |
1 + 1 · x1 + 2 · x2 + 1 · x3
|
all of the following rules can be deleted.
1.1.1 Rule Removal
Using the
prec(a__f) |
= |
1 |
|
stat(a__f) |
= |
mul
|
prec(a) |
= |
2 |
|
stat(a) |
= |
mul
|
prec(a__b) |
= |
2 |
|
stat(a__b) |
= |
mul
|
prec(b) |
= |
0 |
|
stat(b) |
= |
mul
|
prec(f) |
= |
0 |
|
stat(f) |
= |
lex
|
π(a__f) |
= |
[1,2,3] |
π(a) |
= |
[] |
π(a__b) |
= |
[] |
π(b) |
= |
[] |
π(f) |
= |
[2,3,1] |
all of the following rules can be deleted.
a__f(X1,X2,X3) |
→ |
f(X1,X2,X3) |
(6) |
1.1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
[a__f(x1, x2, x3)] |
= |
2 · x1 + 1 · x2 + 2 · x3
|
[a] |
= |
1 |
[a__b] |
= |
1 |
[b] |
= |
0 |
all of the following rules can be deleted.
a__f(a,X,X) |
→ |
a__f(X,a__b,b) |
(1) |
1.1.1.1.1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(a__b) |
= |
1 |
|
weight(a__b) |
= |
1 |
|
|
|
prec(a) |
= |
0 |
|
weight(a) |
= |
1 |
|
|
|
all of the following rules can be deleted.
1.1.1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.