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Unformatted text preview: , and vA : A → B and vB : B → B are valuations, then
vB is called an extension of the valuation vA if and only if vB coincides with
vA when restricted to A.
The consistency conditions V1V6 are quite stringent, as the next theorem
shows. 19 Uniqueness of Valuations
Theorem 1. If two valuations v and v coincide on the set S of symbols, then
they coincide on the set Prop of all propositions.
Proof. Seeking a contradiction, we assume that there exist two valuations v and
v that coincide on S, but do not coincide on Prop. Thus, the set
C = {a ∈ Prop  v a = v a} of counter examples is not empty. Choose a counter example a in C of minimal
length, where the length of the proposition is deﬁned as the number of terminal
symbols. Then a cannot be of the form a = ¬b, since the minimality of the
counter example implies that v b = v b, which implies
v a = M¬ (v b) = M¬ (v b) = v a . Similarly, a cannot be of the form bXc for some propositions b and c in Prop
and some connective X in {∧, ∨, ⊕, →, ↔}. Indeed, by the minimality of the
counter example v b = v b and v c = v c, which implies
v a = MX (v b , v c) = MX (v b , v c) = v a . 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. 20 and some connective X in {∧, ∨, ⊕, →, ↔}. Indeed, by the minimality of the
counter example v b = v b and v c = v c, which implies Interpretation v a = MX (v b , v c) = MX (v b , v c) = v a . 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.
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.
21 Interlude: Induction 22 Strong Induction
Suppose we wish to prove a certain assertion concerning
nonnegative integers.
Let A(n) be the assertion...
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 Fall '11
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