Certification Problem
Input (TPDB TRS_Standard/AProVE_04/Liveness6.2)
The rewrite relation of the following TRS is considered.
|
top(sent(x)) |
→ |
top(check(rest(x))) |
(1) |
|
rest(nil) |
→ |
sent(nil) |
(2) |
|
rest(cons(x,y)) |
→ |
sent(y) |
(3) |
|
check(sent(x)) |
→ |
sent(check(x)) |
(4) |
|
check(rest(x)) |
→ |
rest(check(x)) |
(5) |
|
check(cons(x,y)) |
→ |
cons(check(x),y) |
(6) |
|
check(cons(x,y)) |
→ |
cons(x,check(y)) |
(7) |
|
check(cons(x,y)) |
→ |
cons(x,y) |
(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
| [top(x1)] |
= |
2 · x1
|
| [sent(x1)] |
= |
2 · x1
|
| [check(x1)] |
= |
2 · x1
|
| [rest(x1)] |
= |
1 · x1
|
| [nil] |
= |
0 |
| [cons(x1, x2)] |
= |
1 + 2 · x1 + 2 · x2
|
all of the following rules can be deleted.
|
rest(cons(x,y)) |
→ |
sent(y) |
(3) |
|
check(cons(x,y)) |
→ |
cons(check(x),y) |
(6) |
|
check(cons(x,y)) |
→ |
cons(x,check(y)) |
(7) |
|
check(cons(x,y)) |
→ |
cons(x,y) |
(8) |
1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
top#(sent(x)) |
→ |
top#(check(rest(x))) |
(9) |
|
top#(sent(x)) |
→ |
check#(rest(x)) |
(10) |
|
top#(sent(x)) |
→ |
rest#(x) |
(11) |
|
check#(sent(x)) |
→ |
check#(x) |
(12) |
|
check#(rest(x)) |
→ |
rest#(check(x)) |
(13) |
|
check#(rest(x)) |
→ |
check#(x) |
(14) |
1.1.1 Dependency Graph Processor
The dependency pairs are split into 2
components.
-
The
1st
component contains the
pair
|
top#(sent(x)) |
→ |
top#(check(rest(x))) |
(9) |
1.1.1.1 Monotonic Reduction Pair Processor with Usable Rules
Using the linear polynomial interpretation over the naturals
| [rest(x1)] |
= |
1 · x1
|
| [nil] |
= |
0 |
| [sent(x1)] |
= |
1 · x1
|
| [check(x1)] |
= |
1 · x1
|
| [top#(x1)] |
= |
1 · x1
|
together with the usable
rules
|
rest(nil) |
→ |
sent(nil) |
(2) |
|
check(sent(x)) |
→ |
sent(check(x)) |
(4) |
|
check(rest(x)) |
→ |
rest(check(x)) |
(5) |
(w.r.t. the implicit argument filter of the reduction pair),
the
rule
could be deleted.
1.1.1.1.1 Reduction Pair Processor
Using the matrix interpretations of dimension 2 with strict dimension 1 over the integers
| [top#(x1)] |
= |
+ · x1
|
| [sent(x1)] |
= |
+ · x1
|
| [check(x1)] |
= |
+ · x1
|
| [rest(x1)] |
= |
+ · x1
|
| [nil] |
= |
|
the
pair
|
top#(sent(x)) |
→ |
top#(check(rest(x))) |
(9) |
could be deleted.
1.1.1.1.1.1 P is empty
There are no pairs anymore.
-
The
2nd
component contains the
pair
|
check#(rest(x)) |
→ |
check#(x) |
(14) |
|
check#(sent(x)) |
→ |
check#(x) |
(12) |
1.1.1.2 Monotonic Reduction Pair Processor with Usable Rules
Using the linear polynomial interpretation over the naturals
| [rest(x1)] |
= |
1 · x1
|
| [sent(x1)] |
= |
1 · x1
|
| [check#(x1)] |
= |
1 · x1
|
having no usable rules (w.r.t. the implicit argument filter of the
reduction pair),
the
rule
could be deleted.
1.1.1.2.1 Size-Change Termination
Using size-change termination in combination with
the subterm criterion
one obtains the following initial size-change graphs.
|
check#(rest(x)) |
→ |
check#(x) |
(14) |
|
| 1 |
> |
1 |
|
check#(sent(x)) |
→ |
check#(x) |
(12) |
|
| 1 |
> |
1 |
As there is no critical graph in the transitive closure, there are no infinite chains.