Certification Problem
Input (TPDB TRS_Standard/Zantema_05/z05)
The rewrite relation of the following TRS is considered.
f(a,f(a,x)) |
→ |
f(c,f(b,x)) |
(1) |
f(b,f(b,x)) |
→ |
f(a,f(c,x)) |
(2) |
f(c,f(c,x)) |
→ |
f(b,f(a,x)) |
(3) |
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) |
c |
is mapped to |
c, |
c1(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(a1(x)) |
→ |
c1(b1(x)) |
(7) |
b1(b1(x)) |
→ |
a1(c1(x)) |
(8) |
c1(c1(x)) |
→ |
b1(a1(x)) |
(9) |
f(a,y1) |
→ |
a1(y1) |
(4) |
f(c,y1) |
→ |
c1(y1) |
(5) |
f(b,y1) |
→ |
b1(y1) |
(6) |
1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
[a1(x1)] |
= |
2 + 2 · x1
|
[c1(x1)] |
= |
2 + 2 · x1
|
[b1(x1)] |
= |
2 + 2 · x1
|
[f(x1, x2)] |
= |
2 + 1 · x1 + 2 · x2
|
[a] |
= |
0 |
[c] |
= |
1 |
[b] |
= |
2 |
all of the following rules can be deleted.
f(c,y1) |
→ |
c1(y1) |
(5) |
f(b,y1) |
→ |
b1(y1) |
(6) |
1.1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
[a1(x1)] |
= |
2 + 2 · x1
|
[c1(x1)] |
= |
2 + 2 · x1
|
[b1(x1)] |
= |
2 + 2 · x1
|
[f(x1, x2)] |
= |
1 + 2 · x1 + 2 · x2
|
[a] |
= |
2 |
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.
a1#(a1(x)) |
→ |
c1#(b1(x)) |
(10) |
a1#(a1(x)) |
→ |
b1#(x) |
(11) |
b1#(b1(x)) |
→ |
a1#(c1(x)) |
(12) |
b1#(b1(x)) |
→ |
c1#(x) |
(13) |
c1#(c1(x)) |
→ |
b1#(a1(x)) |
(14) |
c1#(c1(x)) |
→ |
a1#(x) |
(15) |
1.1.1.1.1 Monotonic Reduction Pair Processor
Using the linear polynomial interpretation over the naturals
[a1(x1)] |
= |
1 + 2 · x1
|
[c1(x1)] |
= |
1 + 2 · x1
|
[b1(x1)] |
= |
1 + 2 · x1
|
[a1#(x1)] |
= |
1 · x1
|
[c1#(x1)] |
= |
1 · x1
|
[b1#(x1)] |
= |
1 · x1
|
the
pairs
a1#(a1(x)) |
→ |
b1#(x) |
(11) |
b1#(b1(x)) |
→ |
c1#(x) |
(13) |
c1#(c1(x)) |
→ |
a1#(x) |
(15) |
and
no rules
could be deleted.
1.1.1.1.1.1 Narrowing Processor
We consider all narrowings of the pair
below position
1
to get the following set of pairs
a1#(a1(b1(x0))) |
→ |
c1#(a1(c1(x0))) |
(16) |
1.1.1.1.1.1.1 Narrowing Processor
We consider all narrowings of the pair
below position
1
to get the following set of pairs
b1#(b1(c1(x0))) |
→ |
a1#(b1(a1(x0))) |
(17) |
1.1.1.1.1.1.1.1 Narrowing Processor
We consider all narrowings of the pair
below position
1
to get the following set of pairs
c1#(c1(a1(x0))) |
→ |
b1#(c1(b1(x0))) |
(18) |
1.1.1.1.1.1.1.1.1 Reduction Pair Processor
Using the matrix interpretations of dimension 3 with strict dimension 1 over the arctic semiring over the naturals
[a1#(x1)] |
= |
+ · x1
|
[a1(x1)] |
= |
+ · x1
|
[b1(x1)] |
= |
+ · x1
|
[c1#(x1)] |
= |
+ · x1
|
[c1(x1)] |
= |
+ · x1
|
[b1#(x1)] |
= |
+ · x1
|
the
pair
a1#(a1(b1(x0))) |
→ |
c1#(a1(c1(x0))) |
(16) |
could be deleted.
1.1.1.1.1.1.1.1.1.1 Dependency Graph Processor
The dependency pairs are split into 0
components.