Certification Problem
Input (TPDB TRS_Standard/Transformed_CSR_04/Ex15_Luc98_GM)
The rewrite relation of the following TRS is considered.
|
a__and(true,X) |
→ |
mark(X) |
(1) |
|
a__and(false,Y) |
→ |
false |
(2) |
|
a__if(true,X,Y) |
→ |
mark(X) |
(3) |
|
a__if(false,X,Y) |
→ |
mark(Y) |
(4) |
|
a__add(0,X) |
→ |
mark(X) |
(5) |
|
a__add(s(X),Y) |
→ |
s(add(X,Y)) |
(6) |
|
a__first(0,X) |
→ |
nil |
(7) |
|
a__first(s(X),cons(Y,Z)) |
→ |
cons(Y,first(X,Z)) |
(8) |
|
a__from(X) |
→ |
cons(X,from(s(X))) |
(9) |
|
mark(and(X1,X2)) |
→ |
a__and(mark(X1),X2) |
(10) |
|
mark(if(X1,X2,X3)) |
→ |
a__if(mark(X1),X2,X3) |
(11) |
|
mark(add(X1,X2)) |
→ |
a__add(mark(X1),X2) |
(12) |
|
mark(first(X1,X2)) |
→ |
a__first(mark(X1),mark(X2)) |
(13) |
|
mark(from(X)) |
→ |
a__from(X) |
(14) |
|
mark(true) |
→ |
true |
(15) |
|
mark(false) |
→ |
false |
(16) |
|
mark(0) |
→ |
0 |
(17) |
|
mark(s(X)) |
→ |
s(X) |
(18) |
|
mark(nil) |
→ |
nil |
(19) |
|
mark(cons(X1,X2)) |
→ |
cons(X1,X2) |
(20) |
|
a__and(X1,X2) |
→ |
and(X1,X2) |
(21) |
|
a__if(X1,X2,X3) |
→ |
if(X1,X2,X3) |
(22) |
|
a__add(X1,X2) |
→ |
add(X1,X2) |
(23) |
|
a__first(X1,X2) |
→ |
first(X1,X2) |
(24) |
|
a__from(X) |
→ |
from(X) |
(25) |
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
| [a__and(x1, x2)] |
= |
2 + 1 · x1 + 2 · x2
|
| [true] |
= |
1 |
| [mark(x1)] |
= |
2 · x1
|
| [false] |
= |
1 |
| [a__if(x1, x2, x3)] |
= |
2 + 2 · x1 + 2 · x2 + 2 · x3
|
| [a__add(x1, x2)] |
= |
2 + 1 · x1 + 2 · x2
|
| [0] |
= |
1 |
| [s(x1)] |
= |
1 · x1
|
| [add(x1, x2)] |
= |
1 + 1 · x1 + 2 · x2
|
| [a__first(x1, x2)] |
= |
2 + 1 · x1 + 2 · x2
|
| [nil] |
= |
1 |
| [cons(x1, x2)] |
= |
1 + 1 · x1 + 1 · x2
|
| [first(x1, x2)] |
= |
1 + 1 · x1 + 2 · x2
|
| [a__from(x1)] |
= |
2 + 2 · x1
|
| [from(x1)] |
= |
1 + 1 · x1
|
| [and(x1, x2)] |
= |
1 + 1 · x1 + 2 · x2
|
| [if(x1, x2, x3)] |
= |
1 + 2 · x1 + 2 · x2 + 2 · x3
|
all of the following rules can be deleted.
|
a__and(true,X) |
→ |
mark(X) |
(1) |
|
a__and(false,Y) |
→ |
false |
(2) |
|
a__if(true,X,Y) |
→ |
mark(X) |
(3) |
|
a__if(false,X,Y) |
→ |
mark(Y) |
(4) |
|
a__add(0,X) |
→ |
mark(X) |
(5) |
|
a__add(s(X),Y) |
→ |
s(add(X,Y)) |
(6) |
|
a__first(0,X) |
→ |
nil |
(7) |
|
a__first(s(X),cons(Y,Z)) |
→ |
cons(Y,first(X,Z)) |
(8) |
|
mark(true) |
→ |
true |
(15) |
|
mark(false) |
→ |
false |
(16) |
|
mark(0) |
→ |
0 |
(17) |
|
mark(nil) |
→ |
nil |
(19) |
|
mark(cons(X1,X2)) |
→ |
cons(X1,X2) |
(20) |
|
a__and(X1,X2) |
→ |
and(X1,X2) |
(21) |
|
a__if(X1,X2,X3) |
→ |
if(X1,X2,X3) |
(22) |
|
a__add(X1,X2) |
→ |
add(X1,X2) |
(23) |
|
a__first(X1,X2) |
→ |
first(X1,X2) |
(24) |
|
a__from(X) |
→ |
from(X) |
(25) |
1.1 Rule Removal
Using the
linear polynomial interpretation over the naturals
| [a__from(x1)] |
= |
2 + 2 · x1
|
| [cons(x1, x2)] |
= |
1 · x1 + 1 · x2
|
| [from(x1)] |
= |
1 · x1
|
| [s(x1)] |
= |
1 + 1 · x1
|
| [mark(x1)] |
= |
2 + 2 · x1
|
| [and(x1, x2)] |
= |
1 + 2 · x1 + 1 · x2
|
| [a__and(x1, x2)] |
= |
1 · x1 + 2 · x2
|
| [if(x1, x2, x3)] |
= |
1 + 2 · x1 + 1 · x2 + 1 · x3
|
| [a__if(x1, x2, x3)] |
= |
1 + 1 · x1 + 2 · x2 + 2 · x3
|
| [add(x1, x2)] |
= |
2 + 2 · x1 + 1 · x2
|
| [a__add(x1, x2)] |
= |
1 + 2 · x1 + 2 · x2
|
| [first(x1, x2)] |
= |
2 + 2 · x1 + 2 · x2
|
| [a__first(x1, x2)] |
= |
1 · x1 + 1 · x2
|
all of the following rules can be deleted.
|
a__from(X) |
→ |
cons(X,from(s(X))) |
(9) |
|
mark(and(X1,X2)) |
→ |
a__and(mark(X1),X2) |
(10) |
|
mark(if(X1,X2,X3)) |
→ |
a__if(mark(X1),X2,X3) |
(11) |
|
mark(add(X1,X2)) |
→ |
a__add(mark(X1),X2) |
(12) |
|
mark(first(X1,X2)) |
→ |
a__first(mark(X1),mark(X2)) |
(13) |
|
mark(s(X)) |
→ |
s(X) |
(18) |
1.1.1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight functions
| prec(mark) |
= |
2 |
|
weight(mark) |
= |
1 |
|
|
|
| prec(from) |
= |
0 |
|
weight(from) |
= |
1 |
|
|
|
| prec(a__from) |
= |
1 |
|
weight(a__from) |
= |
2 |
|
|
|
all of the following rules can be deleted.
|
mark(from(X)) |
→ |
a__from(X) |
(14) |
1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.