Certification Problem
Input (TPDB TRS_Standard/Beerendonk_07/7)
The rewrite relation of the following TRS is considered.
|
cond(true,x) |
→ |
cond(odd(x),p(p(p(x)))) |
(1) |
|
odd(0) |
→ |
false |
(2) |
|
odd(s(0)) |
→ |
true |
(3) |
|
odd(s(s(x))) |
→ |
odd(x) |
(4) |
|
p(0) |
→ |
0 |
(5) |
|
p(s(x)) |
→ |
x |
(6) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by ttt2 @ termCOMP 2023)
1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
cond#(true,x) |
→ |
p#(x) |
(7) |
|
cond#(true,x) |
→ |
p#(p(x)) |
(8) |
|
cond#(true,x) |
→ |
p#(p(p(x))) |
(9) |
|
cond#(true,x) |
→ |
odd#(x) |
(10) |
|
cond#(true,x) |
→ |
cond#(odd(x),p(p(p(x)))) |
(11) |
|
odd#(s(s(x))) |
→ |
odd#(x) |
(12) |
1.1 Dependency Graph Processor
The dependency pairs are split into 2
components.
-
The
1st
component contains the
pair
|
cond#(true,x) |
→ |
cond#(odd(x),p(p(p(x)))) |
(11) |
1.1.1 Reduction Pair Processor with Usable Rules
Using the linear polynomial interpretation over the arctic semiring over the integers
| [0] |
= |
0 |
| [p(x1)] |
= |
-1 · x1 + 0 |
| [false] |
= |
0 |
| [true] |
= |
2 |
| [cond#(x1, x2)] |
= |
4 · x1 + 5 · x2 + 0 |
| [s(x1)] |
= |
2 · x1 +
-∞ |
| [odd(x1)] |
= |
0 · x1 +
-∞ |
together with the usable
rules
|
p(0) |
→ |
0 |
(5) |
|
p(s(x)) |
→ |
x |
(6) |
|
odd(0) |
→ |
false |
(2) |
|
odd(s(0)) |
→ |
true |
(3) |
|
odd(s(s(x))) |
→ |
odd(x) |
(4) |
(w.r.t. the implicit argument filter of the reduction pair),
the
pair
|
cond#(true,x) |
→ |
cond#(odd(x),p(p(p(x)))) |
(11) |
could be deleted.
1.1.1.1 P is empty
There are no pairs anymore.
-
The
2nd
component contains the
pair
|
odd#(s(s(x))) |
→ |
odd#(x) |
(12) |
1.1.2 Size-Change Termination
Using size-change termination in combination with
the subterm criterion
one obtains the following initial size-change graphs.
|
odd#(s(s(x))) |
→ |
odd#(x) |
(12) |
|
| 1 |
> |
1 |
As there is no critical graph in the transitive closure, there are no infinite chains.