Certification Problem
Input (TPDB SRS_Standard/Zantema_04/z068)
The rewrite relation of the following TRS is considered.
|
C(x1) |
→ |
c(x1) |
(1) |
|
c(c(x1)) |
→ |
x1 |
(2) |
|
b(b(x1)) |
→ |
B(x1) |
(3) |
|
B(B(x1)) |
→ |
b(x1) |
(4) |
|
c(B(c(b(c(x1))))) |
→ |
B(c(b(c(B(c(b(x1))))))) |
(5) |
|
b(B(x1)) |
→ |
x1 |
(6) |
|
B(b(x1)) |
→ |
x1 |
(7) |
|
c(C(x1)) |
→ |
x1 |
(8) |
|
C(c(x1)) |
→ |
x1 |
(9) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 String Reversal
Since only unary symbols occur, one can reverse all terms and obtains the TRS
|
C(x1) |
→ |
c(x1) |
(1) |
|
c(c(x1)) |
→ |
x1 |
(2) |
|
b(b(x1)) |
→ |
B(x1) |
(3) |
|
B(B(x1)) |
→ |
b(x1) |
(4) |
|
c(b(c(B(c(x1))))) |
→ |
b(c(B(c(b(c(B(x1))))))) |
(10) |
|
B(b(x1)) |
→ |
x1 |
(7) |
|
b(B(x1)) |
→ |
x1 |
(6) |
|
C(c(x1)) |
→ |
x1 |
(9) |
|
c(C(x1)) |
→ |
x1 |
(8) |
1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [C(x1)] |
= |
1 · x1 + 1 |
| [c(x1)] |
= |
1 · x1
|
| [b(x1)] |
= |
1 · x1
|
| [B(x1)] |
= |
1 · x1
|
all of the following rules can be deleted.
|
C(x1) |
→ |
c(x1) |
(1) |
|
C(c(x1)) |
→ |
x1 |
(9) |
|
c(C(x1)) |
→ |
x1 |
(8) |
1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [c(x1)] |
= |
1 · x1 + 1 |
| [b(x1)] |
= |
1 · x1
|
| [B(x1)] |
= |
1 · x1
|
all of the following rules can be deleted.
1.1.1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
b#(b(x1)) |
→ |
B#(x1) |
(11) |
|
B#(B(x1)) |
→ |
b#(x1) |
(12) |
|
c#(b(c(B(c(x1))))) |
→ |
b#(c(B(c(b(c(B(x1))))))) |
(13) |
|
c#(b(c(B(c(x1))))) |
→ |
c#(B(c(b(c(B(x1)))))) |
(14) |
|
c#(b(c(B(c(x1))))) |
→ |
B#(c(b(c(B(x1))))) |
(15) |
|
c#(b(c(B(c(x1))))) |
→ |
c#(b(c(B(x1)))) |
(16) |
|
c#(b(c(B(c(x1))))) |
→ |
b#(c(B(x1))) |
(17) |
|
c#(b(c(B(c(x1))))) |
→ |
c#(B(x1)) |
(18) |
|
c#(b(c(B(c(x1))))) |
→ |
B#(x1) |
(19) |
1.1.1.1.1 Dependency Graph Processor
The dependency pairs are split into 2
components.
-
The
1st
component contains the
pair
|
c#(b(c(B(c(x1))))) |
→ |
c#(b(c(B(x1)))) |
(16) |
|
c#(b(c(B(c(x1))))) |
→ |
c#(B(c(b(c(B(x1)))))) |
(14) |
|
c#(b(c(B(c(x1))))) |
→ |
c#(B(x1)) |
(18) |
1.1.1.1.1.1 Reduction Pair Processor
Using the linear polynomial interpretation over the naturals
| [c#(x1)] |
= |
1 · x1
|
| [b(x1)] |
= |
1 · x1
|
| [c(x1)] |
= |
1 + 1 · x1
|
| [B(x1)] |
= |
1 · x1
|
the
pairs
|
c#(b(c(B(c(x1))))) |
→ |
c#(b(c(B(x1)))) |
(16) |
|
c#(b(c(B(c(x1))))) |
→ |
c#(B(x1)) |
(18) |
could be deleted.
1.1.1.1.1.1.1 Reduction Pair Processor
Using the matrix interpretations of dimension 3 with strict dimension 1 over the arctic semiring over the naturals
| [c#(x1)] |
= |
+
|
| 0 |
-∞ |
-∞ |
|
-∞ |
-∞ |
-∞ |
|
-∞ |
-∞ |
-∞ |
|
|
· x1
|
| [b(x1)] |
= |
+ · x1
|
| [c(x1)] |
= |
+ · x1
|
| [B(x1)] |
= |
+ · x1
|
the
pair
|
c#(b(c(B(c(x1))))) |
→ |
c#(B(c(b(c(B(x1)))))) |
(14) |
could be deleted.
1.1.1.1.1.1.1.1 P is empty
There are no pairs anymore.
-
The
2nd
component contains the
pair
|
B#(B(x1)) |
→ |
b#(x1) |
(12) |
|
b#(b(x1)) |
→ |
B#(x1) |
(11) |
1.1.1.1.1.2 Monotonic Reduction Pair Processor with Usable Rules
Using the linear polynomial interpretation over the naturals
| [B(x1)] |
= |
1 · x1
|
| [b(x1)] |
= |
1 · x1
|
| [b#(x1)] |
= |
1 · x1
|
| [B#(x1)] |
= |
1 · x1
|
having no usable rules (w.r.t. the implicit argument filter of the
reduction pair),
the
rule
could be deleted.
1.1.1.1.1.2.1 Size-Change Termination
Using size-change termination in combination with
the subterm criterion
one obtains the following initial size-change graphs.
|
b#(b(x1)) |
→ |
B#(x1) |
(11) |
|
| 1 |
> |
1 |
|
B#(B(x1)) |
→ |
b#(x1) |
(12) |
|
| 1 |
> |
1 |
As there is no critical graph in the transitive closure, there are no infinite chains.