matemática básica (ing. civil)

8/15/2019 Matemática Básica (Ing. Civil)

http://slidepdf.com/reader/full/matematica-basica-ing-civil 1/105



∧∨,, →, ∼, ↔

R2



P (A)

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Capıtulo1

p

q

r

. . .



⇒

⇒

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p

q

r

p :

q :

r :

s :

[email protected]



∧ ∨ → ↔ ∼

p q

∧

∧

y

p ∧ q p q

p q p ∧ q

∧

p q ≡ p q ≡ p q ≡ p q ≡ p q

≡ p q ≡ p q ≡ p q ≡ p

q ≡ p q ≡ p q ≡ p

q

p q

q

p

∨

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p q ∨

p ∨ q

p q

p q p ∨ q

∨

p q ≡ p q ≡ p q ≡ p q ≡ p q ≡ p

q ≡ p q

p q p

q

p q p q

p q ≡ p q ≡ p q ≡ p q

≡ p q ≡ p q ≡

p

q

∼

p

∼ p

p

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p ∼ p

∼

p ≡ p ≡

p ≡ p ≡

p

≡ p ≡

p ≡ p ≡

p

≡ p ≡ p

→

p

q

→

p → q

p q

p q p → q

→

p

q ≡ q

p ≡ p

q ≡ p

q ≡

q p ≡ p q ≡ q ∼ p≡ q

p≡ p q ≡ q p ≡ q p ≡q

p ≡ q

p ≡ q

p ≡ q

p ≡ q

p ≡ q

p

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q p

→

p

q ≡

q

p

→

∧

→

↔

( p → q ) ∧ (q → p)

p ↔ q

p

q

p q p ↔ q

↔

p

q ≡ p

q ≡

p

q ≡ p

q

V ( p) = F V ( p) = V

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p

q

V ( p) = F V (q ) = V

s

s ≡ p ∨ q

V (s) ≡ F ∨ V ≡ V

s

20 %

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p q

s

p

q

s

s

p

q

s → ( p ∧ q )

p

q

s

∼ ( p ∧ q ) → (s∨ ∼ p)

p

q

s

p

q

s ≡

s

q

p

s → (

→ p)

p q s

(q → p) ∼ r

p

q

p

s

≡

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≡ p q s p

≡ ( p ↔ q ) ∨ (s → p)

( p →∼ q )∨(∼ r → s)

(∼ p∧ ∼ q )∨ ∼ q

[(∼ r ∨ q ) ∧ q ] ↔ [(∼ q ∨ r) ∧ s

( p → s) → [( p ∨ q )∧ ∼ q ]

( p →∼ q ) ∨ (∼ r → s) ≡ F

p →∼ q ≡ F ∼ r → s ≡ F

p ≡ F

q ≡ V

r ≡ F

s ≡ F

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90 %

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a

x

20 %

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(3 + 54 = 8) ∧ (5 − 3 = 2) (1 < 2) ∨ (5 + 9 = 14)

(4 + 6 = 9) ↔ (5 − 2 = 4)

p ∧ q q → t

∼ (q ∨ r) ∨ ( p ∨ q )

( p

∨ ∼ q )

→ (

∼ r

∧q )

[( p ∧ q ) ∨ (q ∧ ∼ r)] ↔ ( p∨ ∼ r)

A = ( p →∼ q ) → (r →∼ s)

q,p,r,s

( p ∨ q ) ↔ (r ∧ s)

r s

[(∼ p∧ ∼ q ) ∨ (r ∧ s)] ∧ p

[∼ ( p ∨ q ) ∧ (r ∨ s)] ∨ (∼ p ∧ q )

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[(∼ r∧ ∼ s) → ( p ∨ r)]∧ ∼ (r ∧ s)

p, q, r, s, t, w

V (∼ w →∼ s) = F

V [( p∧ ∼ r) ↔ (s → t)] = V ;

[( p ∧ q ) ∨ r] → s

(s ↔∼ w) → (r∨ ∼ p)

[t → (w∨ ∼ p)]∨ ∼ ( p → r)

p q p ∧ q ∼ p p ∨ q p q p → q p ↔ q

n =

2n =

[∼ ( p ∧ q )] ↔ [∼ p∨ ∼q ]

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n = 2 22 = 4 4

p q [∼ ( p ∧ q )] ↔ [∼ p ∨ ∼ p

[( p ∨ q ) ∧ (∼ q ∨ r)] ∧ (∼ r → p)

n = 3

23 = 8

8

[(∼ p ∨ q ) → r] ↔ [( p∧ ∼ q ) ∨ r]

n = 3 23 = 8 8

[( p ∧ q ) → p] → [(q ∨ r) ∧ (∼ q ∧ ∼ r)]

n = 3

23

= 8

8

p :

q :

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r :

s :

[( p ∧ q ) → r] ∨ [(∼ q ∧ s) → r]

T

C

[∼ ( p∨q )∧ ∼ p] ↔ ( p → q )

n = 2

22 = 4

4

p q [

∼ ( p

∨q )

∧ ∼ p

↔ p

→ q

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[( p → q ) ∨ (r → q )] → ( p → r)

∼ [∼ p →∼ (∼ q ∧ ∼ p)]∨ ∼ (∼ p∨ ∼ q )

[( ∨ ∼ q )∧ ∼ p] ∼ (∼ q → p)

[ p → (q → r)] ↔ [( p∧ ∼ r) →∼ q ]

[( ∼ q )∧ ∼ (r ∧ q )] ↔∼ [( p ∼ q ) → (q ∧ r)]

[(∼ p ∧ r) → q ] ↔ [∼ q ↔ p ∨ r] ( p ↔ q ) (q ∨ ∼ r)

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≡

p → p p ↔ p

∼ ( p∧ ∼ p)

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∼ (∼ p) ≡ p

p ∧ p ≡ p

p∨

p ≡

p

p ∧ q ≡ q ∧ p

p ∨ q ≡ q ∨ p

p ↔ q ≡ q ↔ p

( p ∧ q ) ∧ r ≡ p ∧ (q ∧ r)

( p ∨ q ) ∨ r ≡ p ∨ (q ∨ r)

( p ↔ q ) ↔ r ≡ p ↔ (q ↔ r)

p ∧ (q ∨ r) ≡ ( p ∧ q ) ∨ ( p ∧ r)

p ∨ (q ∧ r) ≡ ( p ∨ q ) ∧ ( p ∨ r)

p → (q ∨ r) ≡ ( p → q ) ∨ ( p → r)

p → (q ∧ r) ≡ ( p → q ) ∧ ( p → r)

∼ ( p ∧ q ) ≡∼ p∨ ∼ q

∼ ( p

∨q )

≡∼ p

∧ ∼ q

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p → q ≡∼ p ∨ q

∼ ( p → q ) ≡∼ q ∧ p

p ↔ q ≡ ( p → q ) ∧ (q → p)

p ↔ q ≡ ( p ∧ q ) ∨ (∼ q ∧ ∼ p)

p ∧ ( p ∨ r) ≡ p

p ∧ (∼ p ∨ r) ≡ p ∧ r

p ∨ ( p ∧ r) ≡ p

p ∨ (∼ p ∧ r) ≡ p ∨ r

p → q ≡∼ q →∼ p

p ↔

q ≡∼

q ↔∼

p

T ∧ T ≡ T

T ∧ p ≡ p

T ∧ C ≡ C

T

∨ p

≡ T

C ∨ C ≡ C

C ∨ p ≡ p

C ∧ p ≡ C

T ∨ C ≡ T

p∧ ∼ p ≡ C

p∨ ∼

p ≡

T

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( p ∨ r) ∨ ( p∨ ∼ r) ∨ ( p ∨ s)

p ∨ r ∨ p∨ ∼ r ∨ p ∨ s

p ∨ p ∨ p ∨ r∨ ∼ r ∨ s

p ∨ r∨ ∼ r ∨ s

p ∨ T ∨ s

T

T

( p

∧ ∼ q )

∨[(

∼ p

∨q )

∧(

∼ p

∨ ∼ q )]

( p∧ ∼ q ) ∨ [(∼ p ∨ q ) ∧ (∼ p∨ ∼ q )]

( p∧ ∼ q ) ∨ [(∼ p ∨ (q ∧ ∼ q )]

( p∧ ∼ q ) ∨ [∼ p ∨ C ]

( p∧ ∼ q )∨ ∼ p

∼ q ∨ ∼ p

∼ q ∨ ∼ p

[∼ ( p ∧ q ) ∧ (∼ p ∨ q )] → (∼ p ∧ q )

[(∼ p∨ ∼ q ) ∧ (∼ p ∨ q )] → (∼ p ∧ q )

[(∼ p ∨ (∼ q ∧ q )] → (∼ p ∧ q )

[(

∼ p

∨C ]

→ (

∼ p

∧q )

∼ p → (∼ p ∧ q )

∼ (∼ p) ∨ (∼ p ∧ q )

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∼ ( p ↔ q ) ⇔∼ p ↔ q

∼ ( p ↔ q ) ⇔ ∼ [( p → q ) ∧ (q → p)]

⇔ ∼ ( p → q )∨ ∼ (q → p)]

⇔ ( p∧ ∼ q ) ∨ (q ∧ ∼ p)]

⇔ [( p∧ ∼ q )∨q ]∧[( p∧ ∼ q )∨ ∼ p]

⇔ [ p ∨ q ] ∧ [∼ q ∨ ∼ p]

⇔ [∼ p → q ] ∧ [q →∼ p]

⇔ ∼ p ↔ q

( p → r) ∨ (q → r) ⇔ ( p ∧ q ) → r

[( p∨ ∼ p)∨q ]∧ [(∼ q ∨r)∧(q ∨ ∼ q )∧ p → [∼ q ∨( p∨ ∼ q )]]

[(∼ p ∨ q ) ∧ (∼ q ∨ ∼ p)] ↔ p

( p → q ) ⇔ (∼ p) ∨ q

( ↔ q ) ⇔ ( p → q ) ∧ (q → p)

∼ ( p → q ) ⇔ p∧ ∼ q

(q →∼ p) ∨ (∼ r →∼ p)

∼ (q →∼ p) ↔ (q ∨ p)

(∼ p∧ ∼ q )∨ ∼ q ↔∼ [( ∨q ) ∧ q ]

∼ ( p → q ) ↔ [( p ∨ q )∧ ∼ q ]

p ↓

q ≡

(∼

p)∧

(∼

q )

∼ ( p ↔ q )

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[(∼ p) ↓ q ] ∨ [q ↓ p]

[(∼ p) ↓ q ] ∨ [(∼ q ) ↓ p]

[(∼

p) ↓

(∼

q )]∨

[q ↓

p]

p ↓ q ≡ (∼ p) ∧ (∼ q )

[( p ↓ q ) ↓ (q ↓ p)] ↔ ( p ∨ q )

∼ ( p ∧ q ) ↔ [ p ↓ q ]

( p ↓ q ) ↔∼ ( p ∨ q )

∼ ( p ↓ q ) ↔ p q

F V

∼ [( p ∧ q ) →∼ r] ∧ (s∧ ∼ s)

∼ [∼ ( p ∧ q ) →∼ q ] ∨ p

[( p → q )∨ ∼ p] ∧ (∼ q → p)

∼ ∼ [∼ (∼ p ∧ q )∧ ∼ q ] → [∼ ( p∨ ∼ q )] [(∼ p∧ ∼ q ) ∨ p ∨ q ] ∧ [( ∧q ) ∨ (∼ p∧ ∼ q ) ∨ p] ∧ (∼ q )

[∼ (∼ p →∼ q ) ↔∼ ( p ∨ q )] ∨ [ p → (∼ p ∧ q ∧ r)]

p1 p2 p3 . . . pn

q

p1 p2 p3 . . . pn

q

p1 p2 p3 . . . pn

q

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∧

p

q

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P → q

∼ q

∴∼ p

[( p → q )∧ ∼ q ] →∼ p

p q [( p

→ q )

∧ ∼ q ]

→ ∼ p

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p

q

p ∨ q

q

∴∼ p

[( p∨

q )∧

q ] →∼

p

p q [( p ∨ q ) ∧ q ] → ∼ p

p ∨ q

∼ p →∼ q ∼ q → r ∼ r

p1 ≡∼ p →∼ q p2 ≡∼ q → r p3 ≡∼ r

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Q ≡ p ∨ q

[(∼ p →∼ q ) ∧ (∼q → r) ∧ (∼ r)] → p ∨ q

n = 3

23 = 8

p q r [(∼ p →∼ q ) ∧ (∼ q → r) ∧ (∼ r)] → p ∨ q

→

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p ↔∼

q

r ∨ q

∼ r

∴∼ q

[( p ↔∼ q ) ∧ (q ∨ r)∧ ∼ r] →∼ q

[( p ↔∼ q ) ∧ (q ∨ r)∧ ∼ r] →∼ q ≡ F

[( p ↔∼ q ) ∧ (q ∨ r)∧ ∼ r] ≡ V

∼ q ≡ F

q ≡ V

p

↔∼ q

≡ V q

∨r

≡ V

∼ r

≡ V

p ≡ V

q ≡ V

r ≡ F

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p

q

s

p → q

∼ p → s

p∨ ∼ p∴ q ∨ s

[( p → q ) ∧ (∼ p → s) ∧ ( p∨ ∼ p)] → (q ∨ s)

[( p → q ) ∧ (∼ p → s) ∧ ( p∨ ∼ p)] → (q ∨ s) ≡ F

q ∨ s ≡ F q ≡ F s ≡ F

q ≡ F p → q ≡ V p ≡ F

s ≡ F ∼ p ≡ V ∼ p → s ≡ V s ≡ V

p∨ ∼ p ≡ V

p ≡ V

F

p → q

p∴ q

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p → q

∼ q

∴∼ p

p ∨ q

∼ p

∴ q

p ↔ q

p

∴ q

p → q

q → r

∴ p → r

p ↔ q

q ↔ r

∴ p ↔ r

p ∧ q

∴ p

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p

∴ p ∨ q

p → (q ∧ ∼ q )

∴∼ p

[( p ∧ q ) ∧ r] → s ∧ [ p ∧ (q ∧ r)] → s

( p ∧ q ∧ r) → s

( p ∧ q ∧ r)

∴ s

p

q

p ∧ q

∴ q

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1) p → q

2) q → r

3) ∼ r

4) p ∨ s/ ∴ s

1) 2)

5) p → r

3)

5)

6) ∼ p

4) 6)

7) s

1) p → q

2) ∼ r →∼ q

3) ∼ (∼ p∧ ∼ t)

4) t → s5) ∼ r/ ∴ s

1) p → q

2) q → r

3) ∼ p → t)

4) t → s5) ∼ r/ ∴ s

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1)

2)

6) p → r

3) ∼ p → t)

4) t → s

5) ∼ r

3)

4)

6) p → r

7) ∼ p → s ≡∼ s → p5) ∼ r

7

6)

8) ∼ s → r

5) ∼ r

8) 5)

s

p

∧ q

p → ∼ q/ ∴∼ q

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( p ∧ q ) → (r ∧ s)

∼ q ∨ ∼ s/ ∴ (∼ p)∨ ∼ q

p ∧ ( p ∨ q )

( p ∨ q ) → r

r → r/ ∴ s

r → ∼ q

p → q

∼ r → s/ ∴ p → s

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p ∧ q

∼ p → q/ ∴∼ q

( p ∧ q ) → (r ∧ s)

∼ q

∨ ∼ s/ ∴ (

∼ p)

∨ ∼ q

p ∧ ( p ∨ q )

( p ∨ q ) → r

r → r/ ∴ s

r → ∼ q

p → q

∼ r → s/ ∴ p → s

n

n2

q q 2 = 3

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∴

∴

∴

p

V

p

p

F

p

[email protected]



p

q

p q

∧

∼ p∧ ∼ q p q

∧ p ∧ q

p q

p

q

∨ ∼ p∨ ∼ q

p

q

∨ p ∨ q

V ≡ 1

F ≡0

( p∨ ∼ q )∨ ∼ r

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∼ r

p∨ ∼ q

∼ r

p

∼ q

[ p ∧ (r∨ ∼ q )] ∨ [q ∧ ∼ r]

q∧ ∼ r

p ∧ (r∨ ∼ q)

q

p

∼ r

r∨ ∼

q

q

p

∼ r

r

∼ q

[( p ∧ q ∧ r) ∨ (∼ p ∧ q ∧ r)]∨ ∼ r

∼ p

p ∼ q

q

∼ p ∼ q

p

q

p

∼ r

∼ p

q

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[(∼ p∧ ∼ q ) ∨ ( p ∨ q )] ∧ p ∨ [q ∧ (∼ p∨ ∼ r)]T ∧ ( p ∨ q ) ∨ (q ∧ ∼ r)

T ∧ [ p ∨ (q ∨ (q ∧ ∼ r))]

T ∧ [ p ∨ q ]

[ p ∨ q ]

p

q

A

[( p∧q )∨ ( p∧r)]∧ [( p∧s)∨( p∧ ∼ s)] B

q ∼ p

∼ p q

∼ q

q

A → B

A

B

A ≡ [( p ∧ q ) ∨ ( p ∧ r)] ∧ [( p ∧ s) ∨ ( p∧ ∼ s)]

A ≡ [ p ∧ (q ∨ r)] ∧ [( p ∧ (s∨ ∼ s)]

A ≡ p ∧ (q ∨ r) ∧ p ∧ T A ≡ p ∧ (q ∨ r)

B

B ≡ [(∼ p ∧ q ) ∨ (∼ p ∧ q )] ∧ (∼ q ∨ q )

B ≡ (∼ p ∧ q ) ∧ T

B

≡ (

∼ p

∧q )

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A → B

A → B ≡ p ∧ (q ∨ r) → (∼ p ∧ q )

A → B ≡ ∼ p ∨ (∼ q ∧ ∼ r) ∨ (∼ p ∧ q )

A → B ≡ ∼ p ∨ (∼ p ∧ q ) ∨ (∼ q ∧ ∼ r)

A → B ≡ ∼ p ∨ (∼ q ∧ ∼ r)

∼ q ∼ r

∼ p

a b

a ∼ b

∼ a ∼ b

(a ∧ b) ∨ (a∧ ∼ b) ∨ (∼ a∧ ∼ b)

a b (a ∧ b) ∨ (a∧ ∼b)

∨ (∼ a∧ ∼ b)

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A ≡ (a ∧ b) ∨ (a∧ ∼ b) ∨ (∼ a∧ ∼ b)

A ≡ [a ∧ (b∨ ∼ b)] ∨ (∼ a∧ ∼ b)

A ≡ a ∨ (∼ a∧ ∼ b)

A ≡ a∨ ∼ b)

a b a∨ ∼ b

a

b

∼ s t

r

∼ s

∼ r

t

∼ t

(∼ s ∧ t) ∧ [(∼ s ∧ t)∨ ∼ r]∧ ∼ t ∨ r

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(∼ s ∧ t)∧ ∼ t ∨ r

(∼ s ∧ C ∨ r

r

D ≡ [( p ∧ r) ∨ (

∧q )] ∧ [( p∧ ∼ s) ∨ ( p ∧ s)]

S ≡

∼ q q

∼ p q

∼ p

q

∼ S → D

D ≡ [( p ∧ r) ∨ ( p ∧ q )] ∧ [( p∧ ∼ s) ∨ ( p ∧ s)]

≡ [ p ∧ (r ∨ q )] ∧ [ p ∧ (∼ s ∨ s)]

≡ [ p ∧ (r ∨ q )] ∧ [ p ∧ T ]

≡ [ p ∧ (r ∨ q )] ∧ p

≡ p

S ≡ [(∼ q ∧ q ) ∨ (∼ p ∧ q )] ∧ [∼ p ∨ q ]

≡ [C ∨ (∼ p ∧ q )] ∧ [∼ p ∨ q ]

≡ ∼ p ∧ q ∧ [∼ p ∨ q ]

≡ ∼ p ∧ q

∼ S → D ≡∼ (∼ p ∧ q ) → p ≡ (∼ p ∧ q ) ∨ p ≡ p ∨ q

∼ [ p →∼ (q ∨ r)]

( p ∨ q ) → [(∼ p ∨ q ) → ( p ∧ q )]

[(r ∨ q ) ∧ p]∨ ∼ r ∧ q

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p

q

∼ q ∼ p

∼ p

∼ p ∼ p

∼ p

q

p

p

p

p q

∼ q

r

∼ q

∼ p ∼ q

p

q

p q

∼ p ∼ q

q

p

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http://sisbib.unmsm.edu.pe/bibvirtualdata/libros/filosofia/intro_logica/1_parte.pdf



Capıtulo2

A , B , C , . . .

a , b , c , d , . . .

. . .

. . .

∈

a A



∈

a ∈ A a

A

b

A

/∈

b /∈ A

b

A

[email protected]



A = a,b,c,d,e,f,g

B = 2, 4, 6, 8, 10, 12, 16

C = 1, 2, 3, 4, 5, 6, 7, 8, 9, . . .

A = x/x

B = x/x > 100

C =

x/x

∅

∅ = x/x = x

U

A = x ∈ /x2 − 1 = 0

A

x = 1

x2 − 1 = 0

B = x B

x

[email protected]



C = ∅ C

∅

A = x ∈ R/x2 + 1 = 0

A

x ∈ R

x2 + 1 = 0

B = x ∈ R/x = ∞

B

∞

C = x ∈ Z/5 < x < 6

C

A = −2, 0, 1, 2

B = 1, 3, 5, 6

C = 0, 2, 4

U = −2, 0, 1, 2, 3, 4, 5, 6

B = 2, 4, 6, 8, 10, 12, 14, 16, . . .

C = 4, 8, 12, 16, 20, 24, 28, 32, . . .

A = 10, 20, 30, 40, 50, 60, . . . U = B

A = x ∈

/0 < x < 8

B = x ∈ Z/ − 1 < x < 7

C = x ∈ / − 5 < x < 0

U = x ∈ Z/ − 1 < x < 8

A B

A = x + 3

B = 2x + 5 x

x = −2

[email protected]



A = xn

B = 2

x

x = n

√ 2

A

B A

B A ⊂ B

A B

A = a,b,c

B = a, b, c

A = x ∈

/x

B = x ∈

/x

B = 1, 2, 3, 4

C = 5, 2, 3, 4

A

A

P (A)

P (A)

2n

n A

∅

A P (A)

A = 1, 2, 3, 4

4

24

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P (A) = ∅, 1, 2, 3, 4, 1, 2, 1, 3, 1, 4, 2, 3, 2, 4, 3, 4,

1, 2, 3, 1, 2, 4, 1, 4, 3, 4, 2, 3, 1, 2, 3, 4

A = x ∈ R/x = x ∧ x = x

x = x ∧ x = x A

A = x ∈

/x2 + 3x + 2 = 0

x2 + 3x + 2 = 0 ⇒ (x + 2)(x + 1) = 0

⇒ x + 2 = 0 ∨ x + 1 = 0

⇒ x = −2 ∨ x = −1

A = x ∈

/x = −2 ∨ x = −1

A

A =

x

∈Q/3 < x < 4

A

x = 3 + 4

2

A = x ∈

/x2 − 1 = 0

x2 − 1 = 0 ⇒ (x − 1)(x + 1) = 0

⇒ x

−1 = 0

∨x + 1 = 0

⇒ x = −1 ∨ x = 1

A = x ∈ /x = −1 ∨ x = 1

A

A

A = x ∈ R/x8 + 10x6 + 35x4 + 50x2 + 24 = 0

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F

G

A G

E

G

E

D

G

C D G

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E D G

A G

E

G

A

C

D

B

E − B

B

−A

C − (A ∪ B)

D − (B ∩ G)

D − (B ∩ G)

(A ∪ G) − (B ∩ G)

(A

−G) (C

−F )

A =

B = x/x ∈ , 18 < x < 27

C = x/x2

− 25 = 0 D = 2x + 1/x ∈

, 2 < x < 6 E =

1

2x+5/x ∈

, 1 < x < 9

A = 3, 5, 7, 9, 11

B = +9, −9 C = x/x3 − 7x + 6 = 0

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D = 1, 2, 23, 25, 27, . . .

A = 7, 10, 15, 22, 31, 42, 55, 70

A

10, 22, 42, 70

36, 45, 54, 63, 72

P (∅)

A = m + n, 8, 2m − 2n + 4

B = x/x = mk, k ∈ Z

C = x/x = nk, k ∈ Z

(B ∪ C )

x ∈ Q/10x2 − 13x − 3 = 0

x ∈

/6 < x < 7

x/x

A ⊂ B B ⊂ A

A = x ∈

|x x2 ≤ 143

B =

x

∈

|x + 1 x

≤ 10

A = x ∈

|x

x2 ≤ 130

B = x ∈

|x + 1

x ≤ 12

A = x ∈

|x x2 ≥ 4 x2 ≤ 141

B = x ∈ |x − 1

x ≤ 9

A = x ∈

|x

x2 ≤ 150

B = x ∈

|x − 1

x ≤ 11

A = x|x x2

B = x|x

x

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A

x ∈ A A − x

A P (A)

P (x)

x c

P (x) : x + 5 > 3

Q(x) : 2x + 5 = 0

R(x) : x

x

∀

x

∀x

x

P (x)

x

P (x)

∀x : P (x)

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P (x) : x ≥ 0

x

∀x ∈ :

P (x) = ∀

x ∈

: x ≥

0

x

∃

x ∃x

x

P (x)

x

P (x)

∃x : P (x)

P (x) : x ≥ 0

x

∃x ∈

:

P (x) = ∃x ∈ : x ≥ 0

x

A = B ⇔ ∀x : x ∈ A ↔ x ∈ B

A ⊂ B ⇔ ∀x : x ∈ A → x ∈ B

A ⊂ A

A ⊂ B ∧ B ⊂ A → A = B

A ⊂ B ∧ B ⊂ C → A ⊂ C

A

= B

⇔ ∄x : x

∈ A

∧x ∈

B

A = B ⇔ (A ⊂ B) ∧ (A ⊃ B)

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P (x) A

∼ [∀x ∈ A : P (x)] ≡ ∃x ∈ A :∼ P (x)

∼ [∃x ∈ A : P (x)] ≡ ∀x ∈ A :∼ P (x)

P (x) : x

x

x

x =

A1, 2, 3, 4, 5

p : ∃x ∈ A/(x + 2 = 6) → (x − 5 = 8)

q : ∀x ∈ A/(x + 2 > 2) ∨ (x + 2 < 2)

r : ∀x ∈ A, ∃y ∈ A/x + y > 2

s =∼ [( p → q ) ∧ (q ∨ ∼ r)]

∃x ∈ R/x2 > 0

∀x ∈ R/x2 ≤ 0

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∀x ∈ /x2 ≥ x

∃x ∈ /x2 < x

∀x ∈ Z/x > 1 → x2 > x

A = −2, −1, 0, 1, 2 U = x ∈ A/x > 2 →x < 2

p

∀x ∈ U/x > 3 ∨ x < 2

q

∃x ∈ U/x2 = 2 → x > 1

s

∀x ∈ U/x2−4

x+2 = x − 2

m,n,t [(∼ p ∧n)∧(t → s)] → (q ∧ m) ≡ F

U

U = x ∈ A/x > 2 → x < 2= x ∈ A/ ∼ (x > 2) ∨ x < 2= x ∈ A/x ≤ 2 ∨ x < 2= x ∈ A/x ≤ 2=

−2,

−1, 0, 1, 2

p : ∀x ∈ U/x > 3 ∨ x < 2 ≡ F

q : ∃x ∈ U/x2 = 2 → x > 1

: ∃x ∈ U/ ∼ (x2 = 2) ∨ x > 1

: ∃x ∈ U/x

2

= 2 ∨ x > 1≡ V

s : ∀x ∈ U/x2 − 4

x + 2 = x − 2

: ∀x ∈ U/x − 2 = x − 2

x = −2

≡ F

m,n,t

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[(∼ p ∧ n) ∧ (t → s)] → (q ∧ m) ≡ F

∼ p ∧ n ≡ V t → s ≡ V (q ∧ m) ≡ F

m ≡

F

t ≡

F

n ≡

V

r n n ≤ r <

n + 1

p : ∀r ∈ Q, ∃n ∈ Z/n ≤ r < n + 1

x

N

x2 < M + 1 x < M

p : ∀x ∈ R, ∃M ∈ Z/x < M → x2 < M + 1

x

y

x y

p : ∀x ∈ R, ∃!y ∈ R/x − y > 0

x

x

x =

p :

q :

r :

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p q r

( p∧

q ) →

r

∃

( p ∧ q ) → r

∄ : ( p ∧ q ) → r

∄ :∼ p∨ ∼ q ∨ r

∃ : ( p ∧ q )∧ ∼ r

y

y

y =

P :

Q :

P

Q

P Q

∀ : P Q

∀ :∼ [(P ∧ Q) ∨ (∼ P ∧ ∼ Q)]

∃ : (P ∧ Q) ∨ (∼ P ∧ ∼ Q)

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x

A) = 1, 2, 3, 4, 5 B = −2, −1, 0, 5, 6,

∀x ∈ A, ∃y ∈ B : x + y < 3

∃!y

∈ B,

∀x

∈ A : x

−y > 1

∀x ∈ B, ∀y ∈ A : x < y → x2 < y2

∃x ∈ A, ∀y ∈ B : (x − y) ∈ A

p = ∀x ∈ A, ∃y ∈ A : (x2 > xy − 52)

q = ∃x ∈ A, ∀y ∈ A :∼ (x + y = 0)

r = ∀x ∈ A, ∀ ∈ A :

x2−y2

x−y = x + y

A = x ∈ Z : −50 ≤ x < 50

( p ∧ q ) ↔∼ (r →∼ p)

a b

a < b

b a

a

n

n > n0 n > a

∀a ∈ R, ∀b ∈ R : ab = 0 ↔ (a = 0 ∨ b = 0)

x b br

(b + 1)r

∀ ∈ A, ∃y ∈ A : [P (x, y) → Q(y)]

∃x ∈ A : ∃y ∈ B : P (x) ∧ Q(y)

∃x ∈ C : ∀y ∈ B, P (x) ∨ [∼ Q(y)]

∼ [∀

x ∈

A : P (x) →

Q(x)]≡ ∃

x ∈

A : P (x)∨ ∼

Q(x)

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∀x ∈ R, ∀y ∈ R : (−x)(−y) = xy → xy > 0

[∃x ∈ : x + 2 = 5] ∧ [∀x ∈

; x2 > x]

x y x y y

x

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M

x

y

x

x

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U

A

C

B

U

B

A

C

A B

A ∪ B

A B

A

∪B =

x

∈ U/x

∈ A

∨x

∈ B

U

A B

A∩B

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A B

A ∩ B = x ∈ U/x ∈ A ∧ x ∈ B

U

A B

A−B

A

B

A

−B =

x

∈ U/x

∈ A

∧x /

∈ B

U

A B

A B

A − B B − A

A B = x ∈ U/(x ∈ A ∧ x /∈ B) ∨ (x /∈ A ∧ x ∈ B)

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U

A B

CA, A, AC

U

A

CA = x ∈ U/x /∈ A

U

A

Ω = 0, 1, 2, 3, 4, 5, 6, 6, 7, 8, 9 A = 0, 2, 4, 6, 8

B =

1, 3, 5, 7, 9

C =

2, 3, 4, 5

D =

1, 6, 7

A ∪ B

A ∩ B

(C ∩ B) ∪ B

A ∩ C ∩ D

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A ⊂ B

A ∪ B = . . .

A ∩ B = . . .

A − B = . . .

A

A = x ∈

/x

1 < x < 10

U =

A =?

A A = 2, 3, 5, 7

A

= − 2, 3, 5, 7

A = x ∈ Z/x = 3n − 1, n ∈ ∧ n < 11

B =

x ∈ Z/x =

5n

2 ,

n ∈

∧ n < 14

A − B

A

B

A = 2, 5, 8, 11, 14, 17, 20, 23, 26, 29B =

5, 10, 15, 20, 25, 60

A − B = 2, 8, 11, 14, 17, 23, 26, 29

A ∪ B = U

B = A

B = A − B

A ∩ B = ∅

(A ∪ B) = ∅

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x

∈ A x /

∈ B x

∈ C

y ∈ A y ∈ B y /∈ C z /∈ A z /∈ B z ∈ C

y /∈ B ∧ z ∈ A

y /∈ B ∧ z ∈ A

V

∧ F

F

x /∈ C ∨ y ∈ B

x /∈ C ∨ y ∈ B

V ∨ V

V

x ∈ A ↔ y ∈ C

(x ∈ A → z ∈ C ) ⇒ (y /∈ A → z ∈ B)

(z ∈ A → x ∈ B) ⇒ y ∈ B

x ⊂ A ↔ y ∈ C

(

x

∈ A

→ z

⊂ C )

⇒ (y /

∈ A

→ z

⊂ B)

(z ⊃ A → x ⊃ B) ⇒ y ∈ B

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(x ∈ A ∧ x ∈ B) ∨ (x ∈ A ∧ x ∈ C )

∩

∪

x ∈ ((A ∩ B) ∪ (A ∩ C ))

A ∩ (B ∪ C ) ⊂ (A ∩ B) ∪ (A ∩ C )

A ∩ (B ∪ C ) ⊃ (A ∩ B) ∪ (A ∩ C )

x ∈ (A ∩ B) ∪ (A ∩ C )

∩

∪ (x ∈ A ∧ x ∈ B) ∨ (x ∈ A ∧ x ∈ C )

x

∈ A

∧(x

∈ B

∨x

∈ C )

∩

∪

x ∈ A ∩ (B ∪ C )

A ∩ (B ∪ C ) ⊃ (A ∩ B) ∪ (A ∩ C )

A ∩ (B ∪ C ) = (A ∩ B) ∪ (A ∩ C )

[(A − B) − C ] ⊂ (A − (B ∪ C ))

x ∈ [(A − B) − C ]

− x ∈ A ∧ x /∈ B ∧ x /∈ C

x ∈ A ∧ x /∈ (B ∪ C )

x ∈ [A − (B ∩ C )]

[(A

−B)

−C ]

⊂ (A

−(B

∪C ))

A B = (A ∩ B ) ∪ (A ∩ B)

A B ⊂ (A ∩ B ) ∪ (A ∩ B)

x ∈ (A B)

x ∈

(A−

B)∨

x ∈

(B −

A)

(x ∈ A ∧ x /∈ B) ∨ (x ∈ B ∧ x /∈ A)

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(x ∈ A ∧ x ∈ B ) ∨ (x ∈ B ∧ x ∈ A)

∩

∪

x ∈ [(A ∩ B ) ∪ (B ∩ A)]

A B ⊂ (A ∩ B ) ∪ (A ∩ B)

A B ⊃ (A ∩ B ) ∪ (A ∩ B)

x ∈ [(A ∩ B ) ∪ (B ∩ A)]

∩

∪

(x ∈ A ∧ x ∈ B ) ∨ (x ∈ B ∧ x ∈ A)

(x

∈ A

∧x /

∈ B)

∨(x

∈ B

∧x /

∈ A)

x ∈ (A − B) ∨ x ∈ (B − A)

x ∈ (A B)

A B ⊃ (A ∩ B ) ∪ (A ∩ B)

A B = (A ∩ B ) ∪ (A ∩ B)

A

B

C

A ⊆ B A ∩ B = A A ∪ B = B

A ∪ A = A A ∩ A = A

A

∪B = B

∪A

A

∩B = B

∩A

(A ∪ B) ∪ C = A ∪ (B ∪ C )

(A ∩ B) ∩ C = A ∩ (B ∩ C )

(A∪B)∪C = (A∪B)∪(A∪C ) (A∩B)∩C = (A∩B)∩(A∩C )

A ⊂ (A ∪ B) A ⊃ (A ∩ B)

A ⊂ B

(A ∪ C ) ⊂ (B ∪ C )

(A ∩ C ) ⊂ (B ∩ C )

A ⊂

B ⇔

A∪

B = B

A ⊂

B ⇔

A∩

B = A

A ∪ A = U A ∩ A = ∅

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A ∪ ∅ = ∅ A ∩ ∅ = ∅

A ∪ U = U

A ∩ U = A

A ∩ (A ∪ B) = A

A ∪ (A ∩ B) = A

A − A = ∅ A − ∅ = A

∅ − A = ∅

(A − B) ⊂ A

A

−B = (A

∪B)

−B = A

−(A

∩B)

B ∩ (A − B) = ∅

A−(B∩C ) = (A−B)∪(A−C )

A−(B∪C ) = (A−B)∩(A−C )

(A − B) − C = A − (B ∪ C )

(A) = A

A−

B = A∩

B

U = ∅

U = ∅

(A ∩ B) = A ∪ B

(A ∪ B) = A ∩ B

A A = ∅

A ∅ = A

A B = B A

(A B) C = A (B C )

(A B) ∩ C = (A ∩ C ) (B ∩ C )

(A B) ∪ (B C ) = (A ∪ B ∪ C ) − (A ∩ B ∩ C )

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P (A)

P (A)

A

P (A) = X : X ⊂ A

P (A)

A

X ∈ P (A) ⇔

X ⊂ A

P (A)

A

A

n

P (A)

2n

∅ A

P (A)

A ⊂ B ⇔ P (A) ⊂ P (B)

X ∈ P (A) ⇒ X ⊂ A ⇒ X ⊂ B ⇒ X ∈ P (B)

x ∈ A ⇒ x ⊂ P (A) ⇒ x ⊂ P (B) ⇒ x ∈ B

P (A) ∪ P (B) ⊂ P (A ∪ B)

X ∈ P (A) ∪ P (B)

X ∈ P (A) ∨ X ∈ P (B) ⇒ x ⊂ (A ∪ B) ⇒ x ∈ P (A ∪ B)

x ∈ A y /∈ A z ∈ A

x ∈ B y /∈ B z /∈ B x /∈ C y ∈ C z /∈ C

x ⊂ A ∧ y /∈ A

z ∈ A x ∈ B

(y /∈ B ↔ z ⊂ B) → [(x ∈ C y ∈ C ) ↔ z ⊂ C

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P (A)

[(x ∈ A → z /∈ B) ∨ (y ∈ B → x ∈ C )] ↔ [(z /∈ A ∧ x /∈ B)]

(z ∈ B ∧ x /∈ B) ∨ y /∈ B ↔ ∼ [(x ∈ C ∨ y /∈ C ) ∧ z ∈ A]

[(x ∈ B y ∈ B) ⇔ (x /∈ C → z ∈ B)] (x /∈ A∨ ∼ z /∈C ) ↔ [∼ (x ∈ A ∧ x ∈ B ∧ x ∈ C ) (y /∈ A ∧ y /∈ B ∧ y /∈C ) (z ∈ A ∧ z ∈ B ∧ z ∈ C ) z ∈ A]

[(x ⊂ A → z /∈ B) ∨ (y ∈ B → x ∈ C )] ↔ [(z /∈ A ∧ x /∈B)]

(z ⊂ B ∧ x /∈ B) ∨ y /∈ B ↔ ∼ [(x ⊂ C ∨ y /∈ C ) ∧ z ∈A]

[(x ∈ B y ∈ B) ⇔ (x /∈ C → z ⊂ B)] (x /∈ A∨ ∼z /∈ C ) ↔ [∼ (x ⊂ A ∧ x ∈ B ∧ x ∈ C ) (y /∈ A ∧ y /∈B ∧ y /∈ C ) (z ∈ A ∧ z ∈ B ∧ z ∈ C ) z ⊂ A]

A ⊂ B ⇔ B ⊂ A

A B = A ⇒ A ⊂ B

(M ⊂ A) ∧ (M ⊂ B) ⇒ M ⊂ A ∩ B

A ⊂ B ⇒ A ∪ (B − A) = A

C (A − B) ⊂ (A − C ) ∪ (C − B)

(A ∪ B ∪ C ) − (A ∩ B ∩ C ) = (A B) ∪ (B C )

A ∩ B = ∅ ⇔ A ⊂ B

A B = B ⇒ B ⊂ A

a = b, c ⇔ a = b = c

M n

n

M 2 ∩ M 3 M 7 ∩ M 8

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P (A)

A ⊂ B ⇒ M ∩ A ⊂ M ∩ B, ∀M

P (A ∩ B) = P (A) ∩ P (B)

A = ∅ P (P (A))

A B

P (A ∪ B) P (A) ∪ P (B)

[(A ∪ B] − (C − A)] ∩ [(A ∩ B) − (A ∩ C )]

[(A ∩ B) ∪ C ] ∪ (B ∪ C )

[C ∪ (B − A)] ∩ [B − (C ∪ A)] ∪ B

A ,B ,C U

[A (B C )] [C B ]

E = ∅, a, a, ∅, a B =

x ∈ E/x = ∅ ∧ x = ∅, a C = x ∈ E/x = a ∧ x = ∅, a

D = x ∈ E /x

P (B − C ) (C ∩ D) − B

M n

n

M 2 ∩ M 3

M 7 ∩ M 8

U = x ∈ /0 < x < 11

A = x ∈U/x

B = x ∈ E/x

C = x ∈ U/x

(A ∪ C ) − B (A − C ) ∩ B

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P (A)

A = ∅ P (P (A))

A B

P (A ∪ B) P (A) ∪ P (B)

A = x ∈

/x3 − 2x2 − 5x + 6 = 0 B =

x ∈

/2x2 − 7x + 3 = 0 C = 2, 3

D = (A − B) ∪ C

P (D)

A = a, ∅, ∅ B = ∅, ∅, ∅

(A ∪ B) − (A ∩ B)

P (A)

P (A) ∩ P (B)

A = x ∈ /7 − x = 3 ∨ x < 3

B = x ∈ /5 − x > 2 ∧ 6x−2

5 ≥ 2

C = x ∈

/x

, x < 10

(A ∪ B) ∩ (C − A) (A − B) ∪ (B ∩ C ) (A B) ∩ (B ∩ C )

A = x ∈

/x = k2−1

2 , k ∈

B = x ∈ /x2 = 8x

C = x ∈ /x2 − 32x + 192 = 0

(B − A) ∩ C

A = Ai/i ∈

Ai

i ∈

B =n

i=1

Ai C =n

i=1

Ai.

A =−1

n, 1

n

/ n ∈

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A = (−n, n) / n ∈

A = (0, n) /

n ∈

A =−n, 1n

/ n ∈

A =

1

2, 1

22, 1

23, . . . ,

1

2n

/ n ∈

∅ = ∅

A − B = B − A

A ⊂ (B ∪ C ) A ⊂ B A ⊂ C

(B ∩ C ) ⊂ A

B ⊂ A C ⊂ A

A n(A)

A ∩ B = ∅ n(A ∪ B) = n(A) +

n(B)

A, B n(A − B) = n(A) − n(A ∩ B)

A, B n(A∪B) = n(A)+n(B)−n(A∩B)

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A A B

B

∼ A = 65 %

A = 35 %

∼ B = 45 % B = 55 %

B ↔ A = 50 %

U

a bc

A B

a + c = 35 % b + c = 55 %

a + b = 50 %

c = 20 %

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T

D

N

a c

g

T D

N

b

de

f

T : a + b + e + d = 620

N : d + e + f + g = 640

D : b + c + e + f = 450

: e = 210: d = 220

: b = 70

: f = 130

a c g

a = 620 − 70 − 210 − 220 = 120

g = 640 − 220 − 210 − 130 = 80

c = 450 − 70 − 210 − 130 = 40

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[(X ∩ Y ) ∪ (Y ∩ Z )] − [X ∩ Y ∩ Z ] = 1, 2, 3, 4

[Z − (X ∪ Y )] ∪ [X ∩ Y ∩ Z ]

[(X ∪ Y ) − (Z ∩ W )] ∪ [(Z − W ) ∩ (Y − X )]

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x + 2y = 87

m + n + o + p + q + r + s = 78 − 10 x + y = 62

x = 37

y = 25

A B C

n(A

B) = n(A)+n(B)−2n(A ∩ B)

n(A ∪ B ∪ C ) = n(A) +

n(B) + n(C ) − n(A ∩ B) −n(B ∩ C ) − n(A ∩ C ) + n(B ∩C ∩ B)

A B C

n[(A B) C ] = n(A) +

n(B) + n(C ) − 2n(A ∩ B) −2n(B ∩ C ) − 2n(A ∩ C ) +

3n(B ∩ C ∩ B)

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72 %

52 %

37 %

32 %

12 %

22 %

2 %

A

B

C

A B

B C A C

A C

A B C

1/4

A

1/5

B

1/3

C

1/2

A B

1/3

B C

1/3

A C

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1/5

1/3

1/4

A B C

A

A B

C

B C

A

C A

C

B

A C

A

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A B

X

A ⊂ X B ⊂ X

A ⊂ Y B ⊂ Y

X ⊂ Y

X = A ∪ B

A, B ⊂ E

A ∩ B = ∅

A ⊂ B

A∪

B = E

A

⊂ B

A, B ⊂ E

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A ⊂ B A ∩B = ∅

A, X ⊂ E

A ∩ X = ∅ A ∪ X = E

X = A

A = B

(A ∩ B ) ∪ (A ∩ B) = ∅

(A

−B)

∪(B

−A) = (A ∪ B) − (A ∩ B).

A B =

A C B = C

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matemática básica (ing. civil)

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