1. Building truth table φ ⇔ ψ

φψφ ⇒ ψψ ⇒ φφ ⇔ ψ
TTTT✔︎
TFFT
FTTF
FFTT✔︎

a. φ ⇔ ψ is true if φ and ψ are both true or both false (φ ⇒ ψ)=(ψ ⇒ φ) , (φ ⇒ ψ)∧(ψ ⇒ φ) = φ ⇔ ψ, φ = ψ

b. φ ⇔ ψ is false if exactly one of φ, ψ is true and the other false.

2. Truth table (φ ⇒ ψ) ⇔ (¬φ ∨ ψ)

φψφ ⇒ ψ¬φ¬φ ∨ ψ
TTTFT
TFFFF
FTTTT
FFTTT

3. Truth table (φψ) ⇔ (φ¬ψ) tautology.

φψφ ⇏ ψ¬ψφ ∧ ¬ψ
TTFFF
TFTTT
FTFFF
FFFTF

4. modus ponens

A. [φ ∧ (φ ⇒ ψ)] ⇒ ψ

| φ | ψ | φ ⇒ ψ | [φ ∧ (φ ⇒ ψ)] | [φ ∧ (φ ⇒ ψ)] ⇒ ψ

TTTTT
TFFFT
FTTFT
FFTFT

b. explain modus ponens

[φ ∧ (φ ⇒ ψ)] ⇒ ψ are all TRUE.

modus ponens: if you know φ and you know φ ⇒ ψ, then you can conclude ψ.

5. Prove Equivalence

by means of a logical argument, in natural language, we can deduce that

¬(φ∧ψ) (¬φ)∨(¬ψ)

  1. (φ∧ψ) , both φ and ψ are true

  2. ¬(φ∧ψ) , not the case that both φ and ψ are true

  3. If they are not both true, then at least one of φ or ψ must be false

  4. the same as saying that, at least one of ¬φ or ¬ψ is true

  5. (¬φ)∨(¬ψ)

  6. This is clearly the same as saying that at least one of ¬φ and ¬ψ is true. (By the definition of negation). here is an error, please change “and” to “or”

¬(φ∨ψ) (¬φ)∧(¬ψ)

  1. (φ∨ψ) , if either φ or ψ is true, then it is true.
  2. ¬(φ∨ψ) , not the case that either φ or ψ is true.
  3. if they are both false, then neither φ nor ψ is true,
  4. the same as saying that, both ¬φ and ¬ψ are true
  5. (¬φ)∧(¬ψ)

6. Denial statement

(a) 34,159 is (a composite number) not a prime number.

(b) Roses are not red or violets are not blue.

(c) If there are no hamburgers, I will have a hot dog.

¬(φ ⇒ ψ) = φ ∧ ¬ ψ, “∧ conjunction uses but / and”

If there are no hamburgers, but I won’t have a hot dog.

(d) Fred will go but he will not play.

this sentence is interpreted as (φ ∧ ¬ ψ)

the denial statement as ¬(φ ∧ ¬ ψ) 1. ¬(φ ∧ ¬ ψ) ⇔ (φ ⇒ ψ),

1.1 if Fred will go, then he will play.

1.2 Fred will go, only if he will play.

1.3 Fred will play if he will go.

1.4 Fred will play whenever he will go.

  1. ¬(φ ∧ ¬ ψ) ⇔ ¬ φ ∨ ψ

2.1 Fred will not go or he will play.

¬(φ ∨ ψ ) ⇔ ¬ φ ∧ ¬ψ

¬(φ ∧ ψ ) ⇔ ¬ φ ∨ ¬ψ

(e) The number x is either negative or greater than 10.

(x<0)∨(x>10) either or 两者任一 disjunction ∨.

¬ [(x<0)∨(x>10)] = (x≥0)∧(x≤10) = 0≤ x ≤10 The number x is non-negative and less than or equal to 10.

(f) We will win the first game or the second.

We will lose the first two games.

7. Show that φ ⇔ ψ = (¬φ) ⇔ (¬ψ).

φφ ⇔ ψψ¬ φ¬ψ(¬φ) ⇔ (¬ψ)
TTTFFT
TTTFFT
FTFTTT
FTFTTT

8. truth tables

(a) φ ⇔ ψ

φφ ⇔ ψψ
TTT
TTT
FTF
FTF

(b) φ⇒(ψ∨θ)

φψψ∨θφ⇒(ψ∨θ)
TTTT
TFTT
FTTT
FFFT

9. [φ ⇒ (ψ ∧ θ)] ⇔ [(φ ⇒ ψ) ∧ (φ ⇒ θ)]

φψθψ∧θφ⇒(ψ∧θ)φ ⇒ ψφ ⇒ θ(φ ⇒ ψ) ∧ (φ ⇒ θ)
TTTTTTTT
TFFFFFFF
FTFFTTTT
FFTFTTTT

10. Verify the equivalence of previous

by means of a logical argument, in natural language, we can deduce that

[φ ⇒ (ψ ∧ θ)] and [(φ ⇒ ψ) ∧ (φ ⇒ θ)]

  1. φ ⇒ (ψ ∧ θ)

  2. ψ and θ from φ

  3. ψ from φ, and θ from φ

  4. (φ ⇒ ψ) and (φ ⇒ θ)

  5. (φ ⇒ ψ) ∧ (φ ⇒ θ)

  6. [φ ⇒ (ψ ∧ θ)] ⇔ [(φ ⇒ ψ) ∧ (φ ⇒ θ)]

c. (φψ )⇒ θ and (φ ⇒ θ) ∧ (ψ ⇒ θ)

  1. φ ∨ ψ ⇒ θ
  2. ****θ from φ or ψ
  3. ****θ from φ or ****θ from ψ
  4. φ ⇒ θ or ψ ⇒ θ
  5. (φ ⇒ θ) ∧ (ψ ⇒ θ)
  6. [(φ ∨ ψ )⇒ θ] ⇔ [(φ ⇒ θ) ∧ (ψ ⇒ θ)]

c0. (φ ⇒ θ) ∧ (ψ ⇒ θ) and (φψ )⇒ θ

  1. (φ ⇒ θ) ∧ (ψ ⇒ θ)
  2. θ from φ or ****θ from ψ
  3. θ from φ or ψ
  4. (φ ∨ ψ )⇒ θ
  5. [ (φ ⇒ θ) ∧ (ψ ⇒ θ)] ⇔ [(φ ∨ ψ )⇒ θ]

11. [φ ⇒ ψ] ⇔ [(¬ψ) ⇒ (¬φ)]

φψφ ⇒ ψ¬ψ¬φ(¬ψ) ⇒ (¬φ)
TTTFFT
TFFTFF
FTTFTT
FFTTTT

(¬ψ) ⇒ (¬φ) is Contrapositives of φ ⇒ ψ.

12. Contrapositives of the statements

(a) rectangles congruent ⇒ same area Contrapositives:

¬ same area ⇒ ¬ rectangles congruent If two rectangles don’t have the same area, they are not congruent.

(b) If a triangle with sides a, b, c (c largest) is right-angled, then $a^2 + b^2$ = $c^2$

Contrapositives:

  1. right-angled ⇒ $a^2 + b^2$ = $c^2$
  2. ¬ ( $a^2 + b^2$ = $c^2$ ) ⇒ ¬ right-angled

If a triangle with sides a, b, c (c largest) and $a^2 + b^2$ ≠ $c^2$, then it is not right-angled.

(c) ( $2^n$ − 1) prime ⇒ $n$ prime

Contrapositives:

¬ $n$ prime ⇒ ¬ ( $2^n$ − 1) prime

If $n$ is not prime, then $2^n$ − 1 is not prime.

(d) Yuan rise ⇒ Dollar fall

Contrapositives:

¬ Dollar fall ⇒ ¬ Yuan rise

If the Dollar not fall, the Yuan will not rises.

13. Converse [φ ⇒ ψ] ⇎ [ψ ⇒ φ]

φψφ ⇒ ψψ ⇒ φ
TTTT
TFFT
FTTF
FFTT

Don’t confuse the contrapositive and converse

14. Converses of the statements 12

(a) rectangles congruent ⇒ same area Converses: rectangles same area ⇒ congruent If two rectangles have the same area, they are congruent. F

(b) triangle ⇒ $a^2 + b^2$ = $c^2$ Converses: ( $a^2 + b^2$ = $c^2$ ) ⇒ triangle If a triangle with sides a, b, c (c largest) and $a^2 + b^2$ = $c^2$, then it is a right-angled. T

(c) ( $2^n$ − 1) prime ⇒ $n$ prime

Converses: $n$ prime ⇒ ( $2^n$ − 1) prime

If $n$ is prime, then $2^n$ − 1 is prime.

(d) Yuan rise ⇒ Dollar fall

Converses: Dollar fall ⇒ Yuan rise

If the Dollar fall, the Yuan will rises.

OPTIONAL PROBLEMS

1. Express the combinator

φ unless ψ

2 ∨ ̇ exclusive or

either one or the other but not both

φψ
TT
TF
FT
FF

3. Express φ ∨ ̇ ψ in terms of the basic combinators ∧, ∨, ¬.

4. true conditional sentence

the converse

φψφ ⇒ ψψ ⇒ φ
FTTF
FFTT

Contrapositives

φψφ ⇒ ψ¬ψ¬φ(¬ψ) ⇒ (¬φ)
TTTFFT
TFFTFF

5. Mod-2 arithmetic

MNM x NM+N
110
100
011
00?0

6. denoting statements

interpret 1 as T and 0 as F

MNM x NM+N
TTF
TF
FT
FF

(a) ∧, ∨ corresponds to × ?

(b) ∧, ∨ corresponds to + ?

(c) Does ¬ correspond to − (minus)?

7. denoting statements

interpret 0 as T and 1 as F

MNM x NM+N
FFT
FTT
TFF
TTT

(a) ∧, ∨ corresponds to × ?

(b) + correspond to ⇒

8 Card puzzle

B, E, 4, 7

at least 3 cards as follows:

A, I, O, U

1, 3, 5, 9,

9. Prove that mn is odd iff m and n are odd.

⇔ iff = if only if = equivalence totally, double direction

mn is odd ⇔ m and n are odd

mn= 15, m=3, n=5

mn= 39, m=3, n=13

True

10. Prove that mn is even iff m and n are even

mn is even ⇔ m and n are even

mn=6, m=2, n=3

False

11. Party

check beer(alcohol) holder’s IDs

12. Compare the logical structure