propositional

# Choose a counter example a in c of minimal length

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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 V1-V6 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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