Certification Problem
Input (TPDB TRS_Standard/Waldmann_06/jwmatchb1)
The rewrite relation of the following TRS is considered.
|
h(f(x,y)) |
→ |
f(y,f(h(h(x)),a)) |
(1) |
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.
|
h#(f(x,y)) |
→ |
h#(h(x)) |
(2) |
|
h#(f(x,y)) |
→ |
h#(x) |
(3) |
1.1.1 Narrowing Processor
We consider all narrowings of the pair
below position
1
to get the following set of pairs
|
h#(f(f(x0,x1),y1)) |
→ |
h#(f(x1,f(h(h(x0)),a))) |
(4) |
1.1.1.1 Forward Instantiation Processor
We instantiate the pair
to the following set of pairs
|
h#(f(f(y_0,y_1),x1)) |
→ |
h#(f(y_0,y_1)) |
(5) |
|
h#(f(f(f(y_0,y_1),y_2),x1)) |
→ |
h#(f(f(y_0,y_1),y_2)) |
(6) |
1.1.1.1.1 Reduction Pair Processor
Using the matrix interpretations of dimension 2 with strict dimension 1 over the integers
| [h#(x1)] |
= |
+ · x1
|
| [f(x1, x2)] |
= |
+ · x1 + · x2
|
| [h(x1)] |
= |
+ · x1
|
| [a] |
= |
|
the
pairs
|
h#(f(f(x0,x1),y1)) |
→ |
h#(f(x1,f(h(h(x0)),a))) |
(4) |
|
h#(f(f(y_0,y_1),x1)) |
→ |
h#(f(y_0,y_1)) |
(5) |
|
h#(f(f(f(y_0,y_1),y_2),x1)) |
→ |
h#(f(f(y_0,y_1),y_2)) |
(6) |
could be deleted.
1.1.1.1.1.1 P is empty
There are no pairs anymore.