∀x ( Human(x) → Mortal(x) )
For every entity x, if x is human, then x is necessarily mortal.
∀x ( Human(x) → Mortal(x) )
For every entity x, if x is human, then x is necessarily mortal.
Human(socrates)
Socrates belongs to the set of human beings.
Mortal(socrates)
It follows, by universal instantiation and modus ponens, that Socrates is mortal.
P → Q
P
∴ Q
If P implies Q, and P holds, then Q holds.
P → Q
¬Q
∴ ¬P
The contrapositive form: deny the consequent, deny the antecedent.
| P | Q | P → Q |
|---|---|---|
| T | T | T |
| T | F | F |
| F | T | T |
| F | F | T |
Material implication: vacuously true when the antecedent is false.
P ∨ Q
¬P
∴ Q
When one disjunct fails, the other must hold.
¬(P ∧ Q)
≡ ¬P ∨ ¬Q
Negation distributes across conjunction by inversion.
P → Q
Q → R
∴ P → R
Implication is transitive — chains of inference compose.
Theorem · √2 ∉ ℚ
The square root of two is irrational. A proof by contradiction, unfolded step by step.
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Assume √2 ∈ ℚ
Suppose, for contradiction, that √2 is rational.
∃ a,b ∈ ℤ : √2 = a/b
Then √2 can be written as a fraction in lowest terms, with a and b coprime integers.
2 = a² / b²
Squaring both sides eliminates the radical.
a² = 2b²
Therefore a² is even, which forces a itself to be even.
a = 2k ⟹ 4k² = 2b²
Let a = 2k. Substituting yields 2k² = b².
b² = 2k²
Hence b² is even, and b is also even.
2 ∣ a ∧ 2 ∣ b
But a and b were assumed coprime — they cannot share the factor 2.
∴ √2 ∉ ℚ
The hypothesis is false. The square root of two is irrational. ∎
論理
quod erat demonstrandum
Which was to be demonstrated.