Certification Problem
Input (TPDB TRS_Standard/SK90/4.28)
The rewrite relation of the following TRS is considered.
|
f(x,nil) |
→ |
g(nil,x) |
(1) |
|
f(x,g(y,z)) |
→ |
g(f(x,y),z) |
(2) |
|
++(x,nil) |
→ |
x |
(3) |
|
++(x,g(y,z)) |
→ |
g(++(x,y),z) |
(4) |
|
null(nil) |
→ |
true |
(5) |
|
null(g(x,y)) |
→ |
false |
(6) |
|
mem(nil,y) |
→ |
false |
(7) |
|
mem(g(x,y),z) |
→ |
or(=(y,z),mem(x,z)) |
(8) |
|
mem(x,max(x)) |
→ |
not(null(x)) |
(9) |
|
max(g(g(nil,x),y)) |
→ |
max'(x,y) |
(10) |
|
max(g(g(g(x,y),z),u)) |
→ |
max'(max(g(g(x,y),z)),u) |
(11) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Rule Removal
Using the
| prec(f) |
= |
4 |
|
stat(f) |
= |
mul
|
| prec(nil) |
= |
0 |
|
stat(nil) |
= |
mul
|
| prec(g) |
= |
3 |
|
stat(g) |
= |
mul
|
| prec(++) |
= |
5 |
|
stat(++) |
= |
lex
|
| prec(null) |
= |
6 |
|
stat(null) |
= |
mul
|
| prec(true) |
= |
6 |
|
stat(true) |
= |
mul
|
| prec(false) |
= |
1 |
|
stat(false) |
= |
mul
|
| prec(mem) |
= |
8 |
|
stat(mem) |
= |
mul
|
| prec(or) |
= |
2 |
|
stat(or) |
= |
lex
|
| prec(=) |
= |
8 |
|
stat(=) |
= |
mul
|
| prec(not) |
= |
7 |
|
stat(not) |
= |
mul
|
| prec(max') |
= |
3 |
|
stat(max') |
= |
mul
|
| prec(u) |
= |
9 |
|
stat(u) |
= |
mul
|
| π(f) |
= |
[1,2] |
| π(nil) |
= |
[] |
| π(g) |
= |
[1,2] |
| π(++) |
= |
[1,2] |
| π(null) |
= |
[1] |
| π(true) |
= |
[] |
| π(false) |
= |
[] |
| π(mem) |
= |
[1,2] |
| π(or) |
= |
[1,2] |
| π(=) |
= |
[1,2] |
| π(max) |
= |
1 |
| π(not) |
= |
[1] |
| π(max') |
= |
[1,2] |
| π(u) |
= |
[] |
all of the following rules can be deleted.
|
f(x,nil) |
→ |
g(nil,x) |
(1) |
|
f(x,g(y,z)) |
→ |
g(f(x,y),z) |
(2) |
|
++(x,nil) |
→ |
x |
(3) |
|
++(x,g(y,z)) |
→ |
g(++(x,y),z) |
(4) |
|
null(nil) |
→ |
true |
(5) |
|
null(g(x,y)) |
→ |
false |
(6) |
|
mem(nil,y) |
→ |
false |
(7) |
|
mem(g(x,y),z) |
→ |
or(=(y,z),mem(x,z)) |
(8) |
|
mem(x,max(x)) |
→ |
not(null(x)) |
(9) |
|
max(g(g(nil,x),y)) |
→ |
max'(x,y) |
(10) |
1.1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight functions
| prec(u) |
= |
1 |
|
weight(u) |
= |
1 |
|
|
|
| prec(max) |
= |
3 |
|
weight(max) |
= |
2 |
|
|
|
| prec(g) |
= |
0 |
|
weight(g) |
= |
0 |
|
|
|
| prec(max') |
= |
2 |
|
weight(max') |
= |
0 |
|
|
|
all of the following rules can be deleted.
|
max(g(g(g(x,y),z),u)) |
→ |
max'(max(g(g(x,y),z)),u) |
(11) |
1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.