Логика. Множества. Вероятность. Лексаченко В.А. - 45 стр.

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R e x e n i e . Dokaem
|
= ((A∧B
)
∨
A ∨ C
)
∼
(A
→
B
∨
(A
→
C))
tret~im sposobom: (
A∧
B
)
∨
A ∨ C
∼ A
→
B
∨
(A
→
C) = (
A
∨
C
) ∼
∼ (A
∨
B
∨
A
∨
C) = (A ∨
C) ∼
((
A∧
C)
∨
A ∨C
) = (A
∨C) ∼ (A
∨C). J
16 .
Dokazat~ sekvenci
A →
B, B → (
A ∼ C
), C
`A
∼ B
tabliqno
(primer 2.2). Preobrazovat~ sekvenci v tavtologi po pra-
vilu vvedeni ∧ i → (upr. 14) i dokazat~ ee: a) s pomow~
sokrawennyh tablic istinnosti, b) s pomow~ teoremy 3.1.
17 . Dokazat~ sekvencii kak v upr. 16.
1)
(
A∧
B
)
→
A
∨ C, A
∨
B ∨ C
|
=
A
→
B
→ (
B
∨ C) ;
2) (A →
B
) →
(
B
∧
C
)
,
(B
→ C) ∨ (
B
→
C)
|
= (
A →
B
) → (B
→
C
) ;
3) A
∨
B ∨
C, C ∨ (
A
∧B
)
|=
A
∨ B
∨
(A
∧
B) ;
4) A ∨ B
∨ C,
(B ∨
C) →
(A
∧B
) |
=
A
∨
(A
∧
B) ;
5)
(A∧
B) →
B, B ∨ (A∧
C)
|
= (
A → B) ∨ (A
∧C
) ;
6)
A
→
(B
→
A), (A → B)
∧
C |= ( A
∨ B
)
→ ((
A → C) → A
) ;
7) C
→
(
A ∨
B
),
C
→ (A∧
B) |= A
∨ B ∨
(A∧B) ;
8)
A →
(A ∨ B
)
, B
→
(A ∨ C) |= (A → B) ∨ (C → A) ;
9)
((A → B)∧
(A →
B
)) →
A,
(A ∨ B)
→ (B∧C
) |= (A →
B
) ∨
B
→ C ;
10) A ∨ (B
∧
C)
,
(
A
→
C)
∧
(
B → C)→(A
∨
B →
C
)
|
=
A →
(
B
→ (A ∨ C)).
R e x e n i e dl A
→
B, B
∨
C |= A
∨
B ∨ C
∨(
A
∧B
∧(
A ∨C
))
. Prime-
n pravilo vvedeni
∧ i
→
(upr. 14), poluqim tavtologi
|=
A → B∧
(
B ∨
C →
A ∨
B ∨ C
∨(A∧
B∧
(
A
∨C
)), kotoru dokaem s
pomow~ teoremy 3.1:
A → B∧B ∨
C →
A ∨
B ∨
C ∨(
A
∧
B∧
(
A
∨C
)) =
= A
∨
B
∨
B ∨
C
∨ A
∨
B ∨
C
∨ (A
∧B
∧(A ∨ C)) = 1
. J
18 .
Dl zadannogo vyvoda napisat~ analiz k kado$i stroke i
postroit~ derevo vyvoda (sm. primer 2.3).
1)
1
. B |
=
A ∨
B 4.
|
=
B
∨ A
→
A
∨
B 7
. A ∨
B
|=
B ∨ A
2. A
|
= A ∨ B 5
. A
|
= B ∨
A
8.
|= A ∨
B
→ B ∨ A
3. B
∨
A |
= A
∨
B
6. B
|
= B
∨ A
9
.
|=
A
∨ B
∼ B
∨
A .
2)
1. A
|
= A
4
.
|
=
A ∨
(
A
∧B) → A 7.
|
= A
∨
(
A
∧
B)
∼
A .
2
. A∧B |=
A 5
. A |=
A ∨ (
A∧
B)
3
. A
∨
(
A∧
B
) |
= A 6.
|
=
A
→ A ∨ (
A∧B
)
3)
1. A∧
(B∧C
) |
=
A
5. A
∧(B
∧
C
) |
=
B
9. A
∧
B, C |= (A
∧
B
)∧C
2. A
∧
(B∧
C) |=
B∧
C
6
. A∧
(B∧
C
)
|=
C
10
. A∧
(
B∧C)
|
= (
A∧B
)
∧
C
3. B
∧C |
=
B 7
. A, B |
=
A∧B
11.
|= A∧(B
∧C) →
4. B∧C
|
=
C 8. A
∧
(
B
∧
C
) |= A∧B
→
(A
∧
B
)
∧
C.
45

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