Unformatted text preview: concerning the integer n.
To prove it for all n >= 0, we can do the following:
1) Prove that the assertion A(0) is true.
2) Assuming that the assertions A(k) are proved for all
k < n, prove that the assertion A(n) is true.
We can conclude that A(n) is true for all n>=0.
23 Example
Theorem: For all n >= 0, we have
1+2+...+ n = n(n+1)/2
Proof. We prove it by strong induction. The assertion A(n) is
the assertion of the theorem.
For n=0, we have 0 = 0(0+1)/2, hence A(0) is true.
Suppose that the assertion A(k) is true for integers 0<=k<n.
Then 1 + 2 +... + n1 + n = (n1)n/2 + n = ((n1)n +2n)/2= (n2+n)/2
= n(n+1)/2. Therefore, A(n) is true.
By the principle of strong induction, A(n) is true for all n>=0.
24 End of Interlude 25 Therefore, a must be a symbol in S, but both valuations coincide on the set S
of symbols, so a cannot be an element of C , which is a contradiction. Degree An interpretation of a proposition p in Prop is an assignment of truth
values to all variables that occur in p. More generally, an interpretation of a set
Y of propositions is an assignment of truth values to all variables that occur in
formulas in Y . The previous theorem states that an interpretation of Prop has
at most one extension to a valuation on Prop.
It remains to show that each interpretation of Prop has an extension to a
valuation. For this purpose, we deﬁne the degree of a proposition p in Prop,
denote deg p, as the number of occurrences of logical connectives in p. In other
words, the degree function satisﬁes the following properties:
D1. An element in S has degree 0.
D2. If a in Prop has degree n, then ¬a has degree n + 1.
D3. If a and b in Prop are respectively of degree na and nb , then aXb is of
degree na + nb + 1 for all connectives X in {∧, ∨, ⊕, →, ↔}.
Example 5. The proposition ((a ∧ b) ∨ ¬c) is of degree 3. Theorem 2. Each interpretation of Prop has a unique extension to a valuation.
Proof. We will show by induction on the degree of a proposition that an interpretation v0 : S → B has an extension to a valuation v : Prop → B. The
uniqueness of this extension is obvious from Theorem 1.
We set v (a) = v0 (a) for all a of degree 0. Then v is certainly a valuation on
the set of degree 0 propositions.
Suppose that v is a valuation for all propositions of degree less than n ex26
tending v0 . If a is a proposition of degree n, then it has a unique formation tree. words, the degree function satisﬁes the following properties:
D1. An element in S has degree 0.
D2. If a in Prop has degree n, then ¬a has degree n + 1.
D3. If a and b in Prop are respectively of degree na and nb , then aXb is of
degree na + nb + 1 for all connectives X in {∧, ∨, ⊕, →, ↔}. Extensions of Interpretations Example 5. The proposition ((a ∧ b) ∨ ¬c) is of degree 3. Theorem 2. Each interpretation of Prop has a unique extension to a valuation.
Proof. We will show by induction on the degree of a proposition that an interpretation v0 : S → B has an extension to a valuation v : Prop → B. The
uniqueness...
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 Fall '11
 math
 Logic, truth values

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