Certification Problem
Input (TPDB TRS_Standard/Zantema_05/z08)
The rewrite relation of the following TRS is considered.
|
f(a,f(b,f(a,x))) |
→ |
f(a,f(b,f(b,f(a,x)))) |
(1) |
|
f(b,f(b,f(b,x))) |
→ |
f(b,f(b,x)) |
(2) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Uncurrying
We uncurry the binary symbol
f
in combination with the following symbol map which also determines the applicative arities of these symbols.
| a |
is mapped to |
a, |
a1(x1) |
| b |
is mapped to |
b, |
b1(x1) |
There are no uncurry rules.
No rules have to be added for the eta-expansion.
Uncurrying the rules and adding the uncurrying rules yields the new set of rules
|
a1(b1(a1(x))) |
→ |
a1(b1(b1(a1(x)))) |
(5) |
|
b1(b1(b1(x))) |
→ |
b1(b1(x)) |
(6) |
|
f(a,y1) |
→ |
a1(y1) |
(3) |
|
f(b,y1) |
→ |
b1(y1) |
(4) |
1.1 Rule Removal
Using the
| prec(f) |
= |
0 |
|
stat(f) |
= |
mul
|
| prec(a) |
= |
1 |
|
stat(a) |
= |
mul
|
| prec(b) |
= |
2 |
|
stat(b) |
= |
mul
|
| π(a1) |
= |
1 |
| π(b1) |
= |
1 |
| π(f) |
= |
[1,2] |
| π(a) |
= |
[] |
| π(b) |
= |
[] |
all of the following rules can be deleted.
|
f(a,y1) |
→ |
a1(y1) |
(3) |
|
f(b,y1) |
→ |
b1(y1) |
(4) |
1.1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
a1#(b1(a1(x))) |
→ |
a1#(b1(b1(a1(x)))) |
(7) |
|
a1#(b1(a1(x))) |
→ |
b1#(b1(a1(x))) |
(8) |
1.1.1.1 Dependency Graph Processor
The dependency pairs are split into 1
component.