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# Logic
TODO Manger at lave en summary.
## Opgaver
### 1.
A) Yes, T
B) Yes, T
C) Yes, T
D) Yes, T
E) No
F) No
### UNKNOWN ORIGIN (12.)
A) if you have the flu, you miss the final examination
B) you pass the course if and only if you do not miss the final examination
C) if you miss the final examination, you do not pass the course
D) you have the flu, miss the final examination and pass the couse
### 15. (13.)
A) ¬p
B) p ^ ¬q
C) p -> q
D) ¬p -> ¬q
E) p -> q
F) q ^ ¬p
G) q -> p
### 20. (18.)
A) T
B) T
C) F
D) T
### UNKNOWN ORIGIN(31.)
A)
| p | ¬p | p ^ ¬p |
| -- | -- | -- |
| T | F | F |
| F | T | F |
B)
| p | ¬p | p v ¬p |
| -- | -- | -- |
| T | F | T |
| F | T | T |
C)
| p | q | p v ¬q | ( p v ¬q) -> q |
| -- | -- | -- | -- |
| T | T | T | T |
| T | F | T | F |
| F | T | F | T |
| F | F | T | F |
D)
| p | q | ( p v q) -> ( p ^ q ) |
| -- | -- | -- |
| T | T | T -> T = T |
| T | F | T -> F = F |
| F | T | T -> F = F |
| F | F | F -> F = T |
E)
| p | q | (p -> q) | (¬q -> ¬p) | (p -> q) <-> (¬q -> ¬p ) |
| -- | -- | -- | -- | -- |
| T | T | T | T | T |
| T | F | F | F | T |
| F | T | T | T | T |
| F | F | T | T | T |
F)
| p | q | (p -> q) | ( q -> p ) | (p -> q) -> ( q -> p) |
| -- | -- | -- | -- | -- |
| T | T | T | T | T |
| T | F | F | T | T |
| F | T | T | F | F |
| F | F | T | T | T |
### (40. )
All the paranterees must be true, they can be split up.
r is in two parantesees ( q v ¬r ) and ( r v ¬p ).
If r is true q must also be true for ( q v ¬p ).
Therefore in ( p v ¬p ) p must also be true.
Therefore it creates a kind of circle, that also works then one is false.
### (44. )
A)
01011
11011
v-----
11011
11000
^-----
11000
B)
01111
10101
^-----
00101
01000
v-----
01101
C)
01010
11011
x-----
10001
01000
x-----
11001
D)
11011
01010
v-----
11011
10001
11011
v-----
11011
11011
11011
^-----
11011
### (7. )
Already done this
### (9. )
The rest is just stupid.
### (20. )
| p | q | p x q | p <-> q |
| -- | -- | -- | -- |
| T | T | F | T |
| T | F | T | F |
| F | T | T | F |
| F | F | F | T |
The two last columns are the negatives of each other thus
p x q = p <-> q
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