Last Update: 20 September 2013
Note: or material is highlighted |
but probably not:
and certainly not:
and t_{i} are terms (noun phrases),
then P(t_{1}, …, t_{n}) is an atomic proposition
Universal Instantiation & Generalization
Existential Instantiation & Generalization
i.e., [(ƒ(a) = b ∧ ƒ(a) = b′) → b = b′]
i.e., same input → same output
i.e., no two objects in range have the same pre-image
i.e., no two outputs have the same input
i.e., (∀a, a′ ∈ A)[ƒ(a) = ƒ(a′) → a = a′]
i.e., if 2 outputs are the same, then their inputs were the same
i.e., no two objects in domain have the same image
i.e., for every possible output, there was an input
i.e., (ƒ o g)(a) = ƒ(g(a))
(∀ set S ⊆ N) [( (0 ∈ S) ∧ (∀k ∈ N) [k ∈ S → k+1 ∈ S] ) → S = N]
(∀ property P) [( P(0) ∧ (∀k ∈ N)[P(k) → P(k+1)]) → (∀n ∈ N)P(n) ]
To prove (∀n ∈ N)P(n):
r² – c_{1}r – c_{2} = 0
i. | R is reflexive | =def | (∀a ∈ A)R(a, a) |
ii. | R is symmetric | =def | (∀a, b ∈ A)[R(a, b) → R(b, a)] |
iii. | R is anti-symmetric | =def | (∀a, b ∈ A)[(R(a, b) ∧ R(b, a)) → a = b] |
iv. | R is transitive | =def | (∀a, b, c ∈ A)[(R(a, b) ∧ R(b, c)) → R(a, c)] |
R is reflexive | ↔ | every vertex has a loop |
R is symmetric | ↔ | for each edge, there is an inverse edge |
R is anti-symmetric | ↔ | the only pairs of inverse edges are loops. |
R is transitive | ↔ | every path of 2 edges has a shortcut. |
Let r be a vertex different from any of these r_{i}.
Then the graph consisting of r with
an edge to each r_{i}
is
a rooted tree with root r.