Certification Problem
Input (TPDB TRS_Innermost/Transformed_CSR_innermost_04/Ex24_Luc06_GM)
The rewrite relation of the following TRS is considered.
|
a__f(b,X,c) |
→ |
a__f(X,a__c,X) |
(1) |
| a__c |
→ |
b |
(2) |
|
mark(f(X1,X2,X3)) |
→ |
a__f(X1,mark(X2),X3) |
(3) |
|
mark(c) |
→ |
a__c |
(4) |
|
mark(b) |
→ |
b |
(5) |
|
a__f(X1,X2,X3) |
→ |
f(X1,X2,X3) |
(6) |
| a__c |
→ |
c |
(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 + 1 · x1 + 2 · x2 + 1 · x3
|
| [b] |
= |
0 |
| [c] |
= |
0 |
| [a__c] |
= |
0 |
| [mark(x1)] |
= |
2 · x1
|
| [f(x1, x2, x3)] |
= |
2 + 1 · x1 + 2 · x2 + 1 · x3
|
all of the following rules can be deleted.
|
mark(f(X1,X2,X3)) |
→ |
a__f(X1,mark(X2),X3) |
(3) |
1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [a__f(x1, x2, x3)] |
= |
1 + 1 · x1 + 2 · x2 + 1 · x3
|
| [b] |
= |
0 |
| [c] |
= |
0 |
| [a__c] |
= |
0 |
| [mark(x1)] |
= |
1 · x1
|
| [f(x1, x2, x3)] |
= |
1 · x1 + 2 · x2 + 1 · x3
|
all of the following rules can be deleted.
|
a__f(X1,X2,X3) |
→ |
f(X1,X2,X3) |
(6) |
1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [a__f(x1, x2, x3)] |
= |
1 · x1 + 2 · x2 + 1 · x3
|
| [b] |
= |
0 |
| [c] |
= |
0 |
| [a__c] |
= |
0 |
| [mark(x1)] |
= |
2 + 2 · x1
|
all of the following rules can be deleted.
|
mark(c) |
→ |
a__c |
(4) |
|
mark(b) |
→ |
b |
(5) |
1.1.1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
a__f#(b,X,c) |
→ |
a__f#(X,a__c,X) |
(8) |
|
a__f#(b,X,c) |
→ |
a__c# |
(9) |
1.1.1.1.1 Dependency Graph Processor
The dependency pairs are split into 0
components.