Certification Problem
Input (TPDB SRS_Standard/Secret_05_SRS/aprove1)
The rewrite relation of the following TRS is considered.
p(0(x1)) |
→ |
0(s(s(p(x1)))) |
(1) |
p(s(x1)) |
→ |
x1 |
(2) |
p(p(s(x1))) |
→ |
p(x1) |
(3) |
f(s(x1)) |
→ |
g(s(x1)) |
(4) |
g(x1) |
→ |
i(s(half(x1))) |
(5) |
i(x1) |
→ |
f(p(x1)) |
(6) |
half(0(x1)) |
→ |
0(s(s(half(x1)))) |
(7) |
half(s(s(x1))) |
→ |
s(half(p(p(s(s(x1)))))) |
(8) |
0(x1) |
→ |
x1 |
(9) |
rd(0(x1)) |
→ |
0(0(0(0(0(0(rd(x1))))))) |
(10) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by ttt2 @ termCOMP 2023)
1 Rule Removal
Using the
linear polynomial interpretation over the naturals
[g(x1)] |
= |
1 · x1 + 17 |
[p(x1)] |
= |
1 · x1 + 0 |
[f(x1)] |
= |
1 · x1 + 17 |
[half(x1)] |
= |
1 · x1 + 0 |
[0(x1)] |
= |
1 · x1 + 8 |
[rd(x1)] |
= |
8 · x1 + 0 |
[i(x1)] |
= |
1 · x1 + 17 |
[s(x1)] |
= |
1 · x1 + 0 |
all of the following rules can be deleted.
0(x1) |
→ |
x1 |
(9) |
rd(0(x1)) |
→ |
0(0(0(0(0(0(rd(x1))))))) |
(10) |
1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
p#(0(x1)) |
→ |
p#(x1) |
(11) |
p#(p(s(x1))) |
→ |
p#(x1) |
(12) |
f#(s(x1)) |
→ |
g#(s(x1)) |
(13) |
g#(x1) |
→ |
half#(x1) |
(14) |
g#(x1) |
→ |
i#(s(half(x1))) |
(15) |
i#(x1) |
→ |
p#(x1) |
(16) |
i#(x1) |
→ |
f#(p(x1)) |
(17) |
half#(0(x1)) |
→ |
half#(x1) |
(18) |
half#(s(s(x1))) |
→ |
p#(s(s(x1))) |
(19) |
half#(s(s(x1))) |
→ |
p#(p(s(s(x1)))) |
(20) |
half#(s(s(x1))) |
→ |
half#(p(p(s(s(x1))))) |
(21) |
1.1.1 Dependency Graph Processor
The dependency pairs are split into 3
components.
-
The
1st
component contains the
pair
i#(x1) |
→ |
f#(p(x1)) |
(17) |
f#(s(x1)) |
→ |
g#(s(x1)) |
(13) |
g#(x1) |
→ |
i#(s(half(x1))) |
(15) |
1.1.1.1 Reduction Pair Processor with Usable Rules
Using the linear polynomial interpretation over the arctic semiring over the integers
[p(x1)] |
= |
-1 · x1 + 0 |
[half(x1)] |
= |
-1 · x1 + 2 |
[g#(x1)] |
= |
1 · x1 + 4 |
[0(x1)] |
= |
-∞ · x1 + 0 |
[i#(x1)] |
= |
1 · x1 + 3 |
[f#(x1)] |
= |
2 · x1 + 3 |
[s(x1)] |
= |
1 · x1 + 3 |
together with the usable
rules
p(0(x1)) |
→ |
0(s(s(p(x1)))) |
(1) |
p(s(x1)) |
→ |
x1 |
(2) |
p(p(s(x1))) |
→ |
p(x1) |
(3) |
half(0(x1)) |
→ |
0(s(s(half(x1)))) |
(7) |
half(s(s(x1))) |
→ |
s(half(p(p(s(s(x1)))))) |
(8) |
(w.r.t. the implicit argument filter of the reduction pair),
the
pair
f#(s(x1)) |
→ |
g#(s(x1)) |
(13) |
could be deleted.
1.1.1.1.1 Dependency Graph Processor
The dependency pairs are split into 0
components.
-
The
2nd
component contains the
pair
half#(s(s(x1))) |
→ |
half#(p(p(s(s(x1))))) |
(21) |
half#(0(x1)) |
→ |
half#(x1) |
(18) |
1.1.1.2 Reduction Pair Processor with Usable Rules
Using the
prec(half#) |
= |
0 |
|
stat(half#) |
= |
lex
|
prec(s) |
= |
0 |
|
stat(s) |
= |
lex
|
prec(p) |
= |
0 |
|
stat(p) |
= |
lex
|
prec(0) |
= |
0 |
|
stat(0) |
= |
lex
|
π(half#) |
= |
1 |
π(s) |
= |
1 |
π(p) |
= |
1 |
π(0) |
= |
[1] |
together with the usable
rules
p(s(x1)) |
→ |
x1 |
(2) |
p(0(x1)) |
→ |
0(s(s(p(x1)))) |
(1) |
p(p(s(x1))) |
→ |
p(x1) |
(3) |
(w.r.t. the implicit argument filter of the reduction pair),
the
pair
half#(0(x1)) |
→ |
half#(x1) |
(18) |
could be deleted.
1.1.1.2.1 Reduction Pair Processor with Usable Rules
Using the linear polynomial interpretation over the arctic semiring over the integers
[p(x1)] |
= |
-1 · x1 + 0 |
[0(x1)] |
= |
-∞ · x1 + 0 |
[half#(x1)] |
= |
1 · x1 + 0 |
[s(x1)] |
= |
1 · x1 + 0 |
together with the usable
rules
p(s(x1)) |
→ |
x1 |
(2) |
p(0(x1)) |
→ |
0(s(s(p(x1)))) |
(1) |
p(p(s(x1))) |
→ |
p(x1) |
(3) |
(w.r.t. the implicit argument filter of the reduction pair),
the
pair
half#(s(s(x1))) |
→ |
half#(p(p(s(s(x1))))) |
(21) |
could be deleted.
1.1.1.2.1.1 P is empty
There are no pairs anymore.
-
The
3rd
component contains the
pair
p#(p(s(x1))) |
→ |
p#(x1) |
(12) |
p#(0(x1)) |
→ |
p#(x1) |
(11) |
1.1.1.3 Size-Change Termination
Using size-change termination in combination with
the subterm criterion
one obtains the following initial size-change graphs.
p#(p(s(x1))) |
→ |
p#(x1) |
(12) |
|
1 |
> |
1 |
p#(0(x1)) |
→ |
p#(x1) |
(11) |
|
1 |
> |
1 |
As there is no critical graph in the transitive closure, there are no infinite chains.