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Last Update: 29 April 2008
Note: |
but probably not:
and certainly not:
P(t1,...,tn) is an atomic proposition
i.e., if tval(premises) = T, then tval(conclusion) = T
Universal Instantiation & Generalization
Existential Instantiation & Generalization
i.e., [ƒ(a)=b ∧ ƒ(a)=b′ → b=b′]
i.e., no two objects in range have the same pre-image
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
(∀ 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 show (∀n ∈ N)P(n):
Let r be a vertex different from any of these ri.
Then the graph consisting of r with an edge to each ri is a rooted tree with root r.
r² c1r c2 = 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)] |
S o R =def {(a,c) ∈ A × C | (∃b ∈ B)[R(a,b) ∧ S(b,c)]
| R is reflexive | ↔ | every vertex has a loop |
| R is symmetric | ↔ | every edge has an "anti-parallel" "companion" |
| R is anti-symmetric | ↔ | the only "anti-parallel pairs" of edges are loops. |
| R is transitive | ↔ | every path of 2 edges has a shortcut. |
Let G be a connected, planar, simple graph
that
divides the plane into a set R of regions.
Then: