Certification Problem
Input (TPDB TRS_Standard/Zantema_05/z11)
The rewrite relation of the following TRS is considered.
|
a(f,0) |
→ |
a(s,0) |
(1) |
|
a(d,0) |
→ |
0 |
(2) |
|
a(d,a(s,x)) |
→ |
a(s,a(s,a(d,a(p,a(s,x))))) |
(3) |
|
a(f,a(s,x)) |
→ |
a(d,a(f,a(p,a(s,x)))) |
(4) |
|
a(p,a(s,x)) |
→ |
x |
(5) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Uncurrying
We uncurry the binary symbol
a
in combination with the following symbol map which also determines the applicative arities of these symbols.
| f |
is mapped to |
f, |
f1(x1) |
| 0 |
is mapped to |
0 |
| s |
is mapped to |
s, |
s1(x1) |
| d |
is mapped to |
d, |
d1(x1) |
| p |
is mapped to |
p, |
p1(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
|
f1(0) |
→ |
s1(0) |
(10) |
|
d1(0) |
→ |
0 |
(11) |
|
d1(s1(x)) |
→ |
s1(s1(d1(p1(s1(x))))) |
(12) |
|
f1(s1(x)) |
→ |
d1(f1(p1(s1(x)))) |
(13) |
|
p1(s1(x)) |
→ |
x |
(14) |
|
a(f,y1) |
→ |
f1(y1) |
(6) |
|
a(s,y1) |
→ |
s1(y1) |
(7) |
|
a(d,y1) |
→ |
d1(y1) |
(8) |
|
a(p,y1) |
→ |
p1(y1) |
(9) |
1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [f1(x1)] |
= |
2 + 1 · x1
|
| [0] |
= |
0 |
| [s1(x1)] |
= |
1 · x1
|
| [d1(x1)] |
= |
1 · x1
|
| [p1(x1)] |
= |
1 · x1
|
| [a(x1, x2)] |
= |
2 · x1 + 2 · x2
|
| [f] |
= |
2 |
| [s] |
= |
0 |
| [d] |
= |
1 |
| [p] |
= |
0 |
all of the following rules can be deleted.
|
f1(0) |
→ |
s1(0) |
(10) |
|
a(f,y1) |
→ |
f1(y1) |
(6) |
|
a(d,y1) |
→ |
d1(y1) |
(8) |
1.1.1 Rule Removal
Using the
| prec(0) |
= |
0 |
|
stat(0) |
= |
mul
|
| prec(a) |
= |
1 |
|
stat(a) |
= |
mul
|
| prec(s) |
= |
2 |
|
stat(s) |
= |
mul
|
| prec(p) |
= |
3 |
|
stat(p) |
= |
mul
|
| π(d1) |
= |
1 |
| π(0) |
= |
[] |
| π(s1) |
= |
1 |
| π(p1) |
= |
1 |
| π(f1) |
= |
1 |
| π(a) |
= |
[1,2] |
| π(s) |
= |
[] |
| π(p) |
= |
[] |
all of the following rules can be deleted.
|
a(s,y1) |
→ |
s1(y1) |
(7) |
|
a(p,y1) |
→ |
p1(y1) |
(9) |
1.1.1.1 Switch to Innermost Termination
The TRS is overlay and locally confluent:
10Hence, it suffices to show innermost termination in the following.
1.1.1.1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
d1#(s1(x)) |
→ |
d1#(p1(s1(x))) |
(15) |
|
d1#(s1(x)) |
→ |
p1#(s1(x)) |
(16) |
|
f1#(s1(x)) |
→ |
d1#(f1(p1(s1(x)))) |
(17) |
|
f1#(s1(x)) |
→ |
f1#(p1(s1(x))) |
(18) |
|
f1#(s1(x)) |
→ |
p1#(s1(x)) |
(19) |
1.1.1.1.1.1 Dependency Graph Processor
The dependency pairs are split into 2
components.
-
The
1st
component contains the
pair
|
f1#(s1(x)) |
→ |
f1#(p1(s1(x))) |
(18) |
1.1.1.1.1.1.1 Usable Rules Processor
We restrict the rewrite rules to the following usable rules of the DP problem.
1.1.1.1.1.1.1.1 Innermost Lhss Removal Processor
We restrict the innermost strategy to the following left hand sides.
1.1.1.1.1.1.1.1.1 Monotonic Reduction Pair Processor
Using the linear polynomial interpretation over the naturals
| [p1(x1)] |
= |
1 · x1
|
| [s1(x1)] |
= |
2 + 1 · x1
|
| [f1#(x1)] |
= |
2 · x1
|
the
rules
|
d1(0) |
→ |
0 |
(11) |
|
d1(s1(x)) |
→ |
s1(s1(d1(p1(s1(x))))) |
(12) |
|
f1(s1(x)) |
→ |
d1(f1(p1(s1(x)))) |
(13) |
|
p1(s1(x)) |
→ |
x |
(14) |
could be deleted.
1.1.1.1.1.1.1.1.1.1 Switch to TRS Processor
We merge the DPs and rules into one TRS.
1.1.1.1.1.1.1.1.1.1.1 Bounds
The given TRS is
match-(raise)-bounded by 1.
This is shown by the following automaton.
-
final states:
{0, 1, 2}
-
transitions:
|
f1#0(0) |
→ |
0 |
|
f1#0(1) |
→ |
0 |
|
f1#0(2) |
→ |
0 |
|
s10(0) |
→ |
1 |
|
s10(1) |
→ |
1 |
|
s10(2) |
→ |
1 |
|
p10(0) |
→ |
2 |
|
p10(1) |
→ |
2 |
|
p10(2) |
→ |
2 |
|
s11(0) |
→ |
4 |
|
p11(4) |
→ |
3 |
|
f1#1(3) |
→ |
0 |
|
s11(1) |
→ |
4 |
|
s11(2) |
→ |
4 |
-
The
2nd
component contains the
pair
|
d1#(s1(x)) |
→ |
d1#(p1(s1(x))) |
(15) |
1.1.1.1.1.1.2 Usable Rules Processor
We restrict the rewrite rules to the following usable rules of the DP problem.
1.1.1.1.1.1.2.1 Innermost Lhss Removal Processor
We restrict the innermost strategy to the following left hand sides.
1.1.1.1.1.1.2.1.1 Monotonic Reduction Pair Processor
Using the linear polynomial interpretation over the naturals
| [p1(x1)] |
= |
1 · x1
|
| [s1(x1)] |
= |
2 + 1 · x1
|
| [d1#(x1)] |
= |
2 · x1
|
the
rules
|
d1(0) |
→ |
0 |
(11) |
|
d1(s1(x)) |
→ |
s1(s1(d1(p1(s1(x))))) |
(12) |
|
f1(s1(x)) |
→ |
d1(f1(p1(s1(x)))) |
(13) |
|
p1(s1(x)) |
→ |
x |
(14) |
could be deleted.
1.1.1.1.1.1.2.1.1.1 Switch to TRS Processor
We merge the DPs and rules into one TRS.
1.1.1.1.1.1.2.1.1.1.1 Bounds
The given TRS is
match-(raise)-bounded by 1.
This is shown by the following automaton.
-
final states:
{0, 1, 2}
-
transitions:
|
d1#0(0) |
→ |
0 |
|
d1#0(1) |
→ |
0 |
|
d1#0(2) |
→ |
0 |
|
s10(0) |
→ |
1 |
|
s10(1) |
→ |
1 |
|
s10(2) |
→ |
1 |
|
p10(0) |
→ |
2 |
|
p10(1) |
→ |
2 |
|
p10(2) |
→ |
2 |
|
s11(0) |
→ |
4 |
|
p11(4) |
→ |
3 |
|
d1#1(3) |
→ |
0 |
|
s11(1) |
→ |
4 |
|
s11(2) |
→ |
4 |