Certification Problem
Input (TPDB TRS_Standard/AG01/#3.54)
The rewrite relation of the following TRS is considered.
|
f(g(x)) |
→ |
g(f(f(x))) |
(1) |
|
f(h(x)) |
→ |
h(g(x)) |
(2) |
|
f'(s(x),y,y) |
→ |
f'(y,x,s(x)) |
(3) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [f(x1)] |
= |
1 · x1
|
| [g(x1)] |
= |
1 · x1
|
| [h(x1)] |
= |
2 · x1
|
| [f'(x1, x2, x3)] |
= |
2 · x1 + 2 · x2 + 1 · x3
|
| [s(x1)] |
= |
1 + 2 · x1
|
all of the following rules can be deleted.
|
f'(s(x),y,y) |
→ |
f'(y,x,s(x)) |
(3) |
1.1 Switch to Innermost Termination
The TRS is overlay and locally confluent:
10Hence, it suffices to show innermost termination in the following.
1.1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
f#(g(x)) |
→ |
f#(f(x)) |
(4) |
|
f#(g(x)) |
→ |
f#(x) |
(5) |
1.1.1.1 Reduction Pair Processor
Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
| prec(g) |
= |
1 |
|
weight(g) |
= |
1 |
|
|
|
| prec(h) |
= |
0 |
|
weight(h) |
= |
1 |
|
|
|
in combination with the following argument filter
| π(f#) |
= |
1 |
| π(g) |
= |
[1] |
| π(f) |
= |
1 |
| π(h) |
= |
[] |
the
pairs
|
f#(g(x)) |
→ |
f#(f(x)) |
(4) |
|
f#(g(x)) |
→ |
f#(x) |
(5) |
could be deleted.
1.1.1.1.1 P is empty
There are no pairs anymore.