一人の言葉Back to notes

17 October 2026

Discrete Math (A)

The abyss of non-continuous world

Term

Term Description
Proposition a declarative sentence, declare a fact (T or F)
Compound Proposition Proposition with logical connectives (the logic operators)
converse 𝑞→𝑝 = converse 𝑝→𝑞
contrapositive ¬𝑞→¬𝑝≡𝑝→𝑞 = contrapositive 𝑝→𝑞
inverse ¬𝑝→¬𝑞 = inverse 𝑝→𝑞
Commutative Law 交換律
Associated Law 結合律
Distributive Law 𝑝∨(𝑞∧𝑟)≡(𝑝∨𝑞)∧(𝑝∨𝑟)
𝑝∧(𝑞∨𝑟)≡(𝑝∧𝑞)∨(𝑝∧𝑟)
Absorption Law 𝑝∨(𝑝∧𝑞)≡𝑝∧(𝑝∨𝑞)≡𝑝
satisfiable any value for any variable can make it T
unsatisfiable no value for any variable can make it T
taotology Always T
contradiction Always F
contingency neither tautology nor contradiction
Predicates a proposition with parameter 𝑥>0
Propositional function generalized proposition / proposition as function with parameter ( P(x) )
Compound Expression propositional logic with predicates 𝑃(3)∧𝑃(−1)

Symbol

Column1 Column2
U Domain
∧ conjunction/and
∨ disjunction/or
¬ not
→ implication / 𝑝→𝑞≡¬𝑝∨𝑞
↔ if and only if / iff / xnor / (𝑝→𝑞)∧(𝑞→𝑝)
∀ for all
∃ exists/for some

Precendence

  1. ∀ ∃
  2. ¬
  3. ∧ ∨
  4. → ⟺

Implication

Expression

𝑝→𝑞 is equivalent to

  1. if p then q
  2. if p, q
  3. q unless ¬𝑝
  4. q if p
  5. q whenever p
  6. q follows from p
  7. p implies q
  8. p only if q
  9. q when p
  10. p is sufficient for q
  11. q is neccessary for p
  12. a neccessary condition for p is q
  13. a sufficient condition for q is p

Equivalent

𝑝→(𝑞∧𝑟)≡(𝑝→𝑞)∧(𝑝→𝑟)
𝑝→(𝑞∨𝑟)≡(𝑝→𝑞)∨(𝑝→𝑟)
(𝑝∨𝑞)→𝑟≡(𝑝→𝑟)∧(𝑞→𝑟) (de morgan)
(𝑝∧𝑞)→𝑟≡(𝑝→𝑟)∨(𝑞→𝑟)

Logically Equivalent

Prop p,q
𝑝↔𝑞 is tautology
written as ≡,⇔

Quantifier / Predicate

Universal

𝑈=ℕ𝑃(𝑥):𝑥>0∀𝑥(𝑃(𝑥))

for all x, P(x)
All natural number is greater than 0

Existential

𝑈=ℕ𝑃(𝑥):𝑥=1∃𝑥(𝑃(𝑥))

for some x, P(x) / there exists

There is an element in natural number and it is 1

Uniqueness (Optional)

∃!𝑥(𝑃(𝑥))=∃𝑥(𝑃(𝑥)∧∀𝑦(𝑃(𝑦)→𝑦=𝑥))

There is a unique x, one and only one

Examples

U = all people
S(x) = student
H(x) = study hard
G(x) = High score

No one studies hard

∀𝑥[¬𝐻(𝑥)]

All students study hard

∀𝑥[𝑆(𝑥)→𝐻(𝑥)]

Some students get high score

∃𝑥[𝑆(𝑥)∧𝐺(𝑥)]

If any student studies hard then he will get a high score

∀𝑥[𝑆(𝑥)∧𝐻(𝑥)→𝐺(𝑥)]

No student gets a high score

¬∃𝑥[𝐺(𝑥)∧𝑆(𝑥)]

Examples 2

  • All lions are fierce
  • Some lions do not drink coffee
  • Some fierce creatures do not drink coffee

logics:
U=creature
𝐿(𝑥) is lion
𝐹(𝑥) is fierce
𝐶(𝑥) drink coffee

  • ∀𝑥(𝐿(𝑥)→𝐹(𝑥))
  • ∃𝑥(𝐿(𝑥)∧¬𝐶(𝑥))
  • ∃𝑥(𝐹(𝑥)∧¬𝐶(𝑥))

Logically Equivalent

  1. Same truth value(iff)
  2. Valid in every domain of discourse

∀𝑥[𝑃(𝑥)∧𝑄(𝑥)]≡∀𝑥𝑃(𝑥)∧∀𝑥𝑄(𝑥)
check

  1. ∀𝑥[𝑃(𝑥)∧𝑄(𝑥)]→∀𝑥𝑃(𝑥)∧∀𝑥𝑄(𝑥)
  2. ∀𝑥𝑃(𝑥)∧∀𝑥𝑄(𝑥)]→∀𝑥[𝑃(𝑥)∧𝑄(𝑥)]

De Morgan for Quantifier

¬∀𝑥𝑃(𝑥)≡∃𝑥¬𝑃(𝑥)∀¬𝑥𝑃(𝑥)≡¬∃𝑥𝑃(𝑥)

The Argument

𝑝→𝑞𝑞→𝑟∴𝑝→𝑟

is equivalent to ((𝑝→𝑞)∧(𝑞→𝑟))→(𝑝→𝑟)

Rule of Inference

Modus Ponens

𝑝→𝑞𝑝∴𝑞

Modus Tollens

𝑝→𝑞¬𝑞∴¬𝑝

Hypothetical Syllogism

𝑝→𝑞𝑞→𝑟∴𝑝→𝑟

Disjunctive Syllogism

𝑝∨𝑞¬𝑝∴𝑞

Addition

𝑝∴𝑝∨𝑞

Simplification

𝑝∧𝑞∴𝑞

Resolution

𝑝∨𝑟¬𝑝∨𝑞∴𝑞∨𝑟

(𝑝∧𝑞)∨(¬𝑝∧𝑟)≡𝑝∧(𝑞∨𝑟)

Universal Instantiation (UI)

∀𝑥𝑃(𝑥)∴𝑃(𝑐)

Universal Generalization (UG)

𝑃(𝑐)for an arbitary c∴∀𝑥𝑃(𝑥)

Existential Instantiation (EI)

∃𝑥𝑃(𝑥)∴𝑃(𝑐)for some element c

Existential Generalization (EG)

𝑃(𝑐)for some c∴∃𝑥𝑃(𝑥)

Valid Argument

example: 𝑝∧(𝑝→𝑞)

StepReason1𝑝∧(𝑝→𝑞)premise2𝑝simplification3𝑝→𝑞simplification4𝑞Modus Ponens from 2,3

Example

∀𝑥[𝑃(𝑥)∨𝑄(𝑥)]∀𝑥[𝑃(𝑥)→𝑅(𝑥)]∴∀𝑥[¬𝑅(𝑥)→𝑄(𝑥)]
StepReason1∀𝑥[𝑃(𝑥)∨𝑄(𝑥)]premise2𝑃(𝑥)∨𝑄(𝑥)UI 13∀𝑥[𝑃(𝑥)→𝑅(𝑥)]premise4𝑃(𝑥)→𝑅(𝑥)UI 35¬𝑅(𝑥)→¬𝑃(𝑥)contrapositive 46𝑅(𝑥)∨(𝑃(𝑥)∨𝑄(𝑥))addition 27𝑅(𝑥)∨¬𝑃(𝑥)equivalent 58𝑅(𝑥)∨¬𝑃(𝑥)∨𝑄(𝑥)addition 79𝑅(𝑥)∨(¬𝑃(𝑥)∧𝑃(𝑥))∨𝑄(𝑥)conj 8 6 with distributive law10𝑅(𝑥)∨𝑄(𝑥)negation law 911¬¬𝑅(𝑥)∨𝑄(𝑥)double negation 1012¬𝑅(𝑥)→𝑄(𝑥)equivalent

Example 2 (28 p84)

∀𝑥[𝑃(𝑥)∨𝑄(𝑥)]∀𝑥[(¬𝑃(𝑥)∧𝑄(𝑥))→𝑅(𝑥)]∴∀𝑥[¬𝑅(𝑥)→𝑃(𝑥)]
StepReason1∀𝑥[𝑃(𝑥)∨𝑄(𝑥)]premise2𝑃(𝑥)∨𝑄(𝑥)UI 13∀𝑥[(¬𝑃(𝑥)∧𝑄(𝑥))→𝑅(𝑥)]premise4(¬𝑃(𝑥)∧𝑄(𝑥))→𝑅(𝑥)UI 35¬𝑅(𝑥)→¬(¬𝑃(𝑥)∧𝑄(𝑥))equivalent6¬𝑅(𝑥)→(𝑃(𝑥)∨¬𝑄(𝑥))equivalent7𝑅(𝑥)∨𝑃(𝑥)∨¬𝑄(𝑥)equivalent8𝑅(𝑥)∨𝑃(𝑥)∨𝑄(𝑥)addition 29𝑅(𝑥)∨𝑃(𝑥)conj 7 810¬𝑅(𝑥)→𝑃(𝑥)equivalent11∀𝑥[¬𝑅(𝑥)→𝑃(𝑥)]UG 10

Tricks

(𝑎∧𝑏)∨(¬𝑎∧𝑐)∨(𝑏∧𝑐)≡(𝑎∧𝑏)∨(¬𝑎∧𝑐)∨(𝑏∧𝑐∧(𝑎∧¬𝑎))≡(𝑎∧𝑏)∨(¬𝑎∧𝑐)∨(𝑏∧𝑐∧𝑎)∨(𝑏∧𝑐∧¬𝑎)≡(𝑎∧𝑏)∨(¬𝑎∧𝑐)∨(𝑏∧𝑐∧¬𝑎)≡(𝑎∧𝑏)∨(¬𝑎∧𝑐)