Certification Problem
Input (TPDB TRS_Standard/SK90/4.24)
The rewrite relation of the following TRS is considered.
|
rev(nil) |
→ |
nil |
(1) |
|
rev(++(x,y)) |
→ |
++(rev1(x,y),rev2(x,y)) |
(2) |
|
rev1(x,nil) |
→ |
x |
(3) |
|
rev1(x,++(y,z)) |
→ |
rev1(y,z) |
(4) |
|
rev2(x,nil) |
→ |
nil |
(5) |
|
rev2(x,++(y,z)) |
→ |
rev(++(x,rev(rev2(y,z)))) |
(6) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Switch to Innermost Termination
The TRS is overlay and locally confluent:
10Hence, it suffices to show innermost termination in the following.
1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
rev#(++(x,y)) |
→ |
rev1#(x,y) |
(7) |
|
rev#(++(x,y)) |
→ |
rev2#(x,y) |
(8) |
|
rev1#(x,++(y,z)) |
→ |
rev1#(y,z) |
(9) |
|
rev2#(x,++(y,z)) |
→ |
rev#(++(x,rev(rev2(y,z)))) |
(10) |
|
rev2#(x,++(y,z)) |
→ |
rev#(rev2(y,z)) |
(11) |
|
rev2#(x,++(y,z)) |
→ |
rev2#(y,z) |
(12) |
1.1.1 Dependency Graph Processor
The dependency pairs are split into 2
components.
-
The
1st
component contains the
pair
|
rev#(++(x,y)) |
→ |
rev2#(x,y) |
(8) |
|
rev2#(x,++(y,z)) |
→ |
rev#(++(x,rev(rev2(y,z)))) |
(10) |
|
rev2#(x,++(y,z)) |
→ |
rev#(rev2(y,z)) |
(11) |
|
rev2#(x,++(y,z)) |
→ |
rev2#(y,z) |
(12) |
1.1.1.1 Reduction Pair Processor with Usable Rules
Using the
| prec(++) |
= |
0 |
|
stat(++) |
= |
lex
|
| prec(nil) |
= |
1 |
|
stat(nil) |
= |
lex
|
| prec(rev1) |
= |
0 |
|
stat(rev1) |
= |
lex
|
| π(rev#) |
= |
1 |
| π(++) |
= |
[2] |
| π(rev2#) |
= |
2 |
| π(rev) |
= |
1 |
| π(rev2) |
= |
2 |
| π(nil) |
= |
[] |
| π(rev1) |
= |
[] |
together with the usable
rules
|
rev2(x,nil) |
→ |
nil |
(5) |
|
rev2(x,++(y,z)) |
→ |
rev(++(x,rev(rev2(y,z)))) |
(6) |
|
rev(nil) |
→ |
nil |
(1) |
|
rev(++(x,y)) |
→ |
++(rev1(x,y),rev2(x,y)) |
(2) |
(w.r.t. the implicit argument filter of the reduction pair),
the
pairs
|
rev#(++(x,y)) |
→ |
rev2#(x,y) |
(8) |
|
rev2#(x,++(y,z)) |
→ |
rev#(rev2(y,z)) |
(11) |
|
rev2#(x,++(y,z)) |
→ |
rev2#(y,z) |
(12) |
could be deleted.
1.1.1.1.1 Reduction Pair Processor with Usable Rules
Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
| prec(rev2#) |
= |
1 |
|
weight(rev2#) |
= |
2 |
|
|
|
| prec(rev#) |
= |
0 |
|
weight(rev#) |
= |
1 |
|
|
|
in combination with the following argument filter
| π(rev2#) |
= |
[] |
| π(rev#) |
= |
[] |
having no usable rules (w.r.t. the implicit argument filter of the
reduction pair),
the
pair
|
rev2#(x,++(y,z)) |
→ |
rev#(++(x,rev(rev2(y,z)))) |
(10) |
could be deleted.
1.1.1.1.1.1 P is empty
There are no pairs anymore.
-
The
2nd
component contains the
pair
|
rev1#(x,++(y,z)) |
→ |
rev1#(y,z) |
(9) |
1.1.1.2 Usable Rules Processor
We restrict the rewrite rules to the following usable rules of the DP problem.
There are no rules.
1.1.1.2.1 Innermost Lhss Removal Processor
We restrict the innermost strategy to the following left hand sides.
There are no lhss.
1.1.1.2.1.1 Size-Change Termination
Using size-change termination in combination with
the subterm criterion
one obtains the following initial size-change graphs.
|
rev1#(x,++(y,z)) |
→ |
rev1#(y,z) |
(9) |
|
|
| 2 |
> |
1 |
| 2 |
> |
2 |
As there is no critical graph in the transitive closure, there are no infinite chains.