Certification Problem
Input (TPDB TRS_Innermost/Mixed_innermost/test830)
The rewrite relation of the following TRS is considered.
|
f(s(X)) |
→ |
f(X) |
(1) |
|
g(cons(0,Y)) |
→ |
g(Y) |
(2) |
|
g(cons(s(X),Y)) |
→ |
s(X) |
(3) |
|
h(cons(X,Y)) |
→ |
h(g(cons(X,Y))) |
(4) |
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
| [f(x1)] |
= |
2 · x1
|
| [s(x1)] |
= |
2 · x1
|
| [g(x1)] |
= |
1 · x1
|
| [cons(x1, x2)] |
= |
1 · x1 + 1 · x2
|
| [0] |
= |
2 |
| [h(x1)] |
= |
2 · x1
|
all of the following rules can be deleted.
1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [f(x1)] |
= |
2 · x1
|
| [s(x1)] |
= |
1 + 1 · x1
|
| [g(x1)] |
= |
1 · x1
|
| [cons(x1, x2)] |
= |
1 · x1 + 2 · x2
|
| [h(x1)] |
= |
1 · x1
|
all of the following rules can be deleted.
1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [g(x1)] |
= |
1 · x1
|
| [cons(x1, x2)] |
= |
1 + 1 · x1 + 1 · x2
|
| [s(x1)] |
= |
1 + 2 · x1
|
| [h(x1)] |
= |
1 · x1
|
all of the following rules can be deleted.
|
g(cons(s(X),Y)) |
→ |
s(X) |
(3) |
1.1.1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
h#(cons(X,Y)) |
→ |
h#(g(cons(X,Y))) |
(5) |
1.1.1.1.1 Usable Rules Processor
We restrict the rewrite rules to the following usable rules of the DP problem.
There are no rules.
1.1.1.1.1.1 Innermost Lhss Removal Processor
We restrict the innermost strategy to the following left hand sides.
|
g(cons(0,x0)) |
|
g(cons(s(x0),x1)) |
1.1.1.1.1.1.1 Reduction Pair Processor
Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
| prec(cons) |
= |
1 |
|
weight(cons) |
= |
1 |
|
|
|
| prec(g) |
= |
0 |
|
weight(g) |
= |
1 |
|
|
|
in combination with the following argument filter
| π(h#) |
= |
1 |
| π(cons) |
= |
[] |
| π(g) |
= |
[] |
the
pair
|
h#(cons(X,Y)) |
→ |
h#(g(cons(X,Y))) |
(5) |
could be deleted.
1.1.1.1.1.1.1.1 P is empty
There are no pairs anymore.