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**Example text**

First, we assume φ is true and have to come up with a proof of χ. 2. Next, we assume ψ is true and need to give a proof of χ as well. 3. Given these two proofs, we can infer χ from the truth of φ ∨ ψ, since our case analysis above is exhaustive. Therefore, we write the rule ∨e as follows: φ∨ψ φ .. ψ .. χ χ χ ∨e. It is saying that: if φ ∨ ψ is true and – no matter whether we assume φ or we assume ψ – we can get a proof of χ, then we are entitled to deduce χ anyway. Let’s look at a proof that p ∨ q q ∨ p is valid: 1 p∨q premise 2 p assumption 3 q∨p ∨i2 2 4 q assumption 5 q∨p ∨i1 4 6 q∨p ∨e 1, 2−3, 4−5 1 Propositional logic 18 Here are some points you need to remember about applying the ∨e rule.

N ψ is valid. Do we have any evidence that these rules are all correct in the sense that valid sequents all ‘preserve truth’ computed by our truth-table semantics? Given a proof of φ1 , φ2 , . . , φn ψ, is it conceivable that there is a valuation in which ψ above is false although all propositions φ1 , φ2 , . . , φn are true? Fortunately, this is not the case and in this subsection we demonstrate why this is so. Let us suppose that some proof in our natural deduction calculus has established that the sequent φ1 , φ2 , .

There is a variant of mathematical induction in which the induction hypothesis for proving M (n + 1) is not just M (n), but the conjunction M (1) ∧ M (2) ∧ · · · ∧ M (n). In that variant, called courseof-values induction, there doesn’t have to be an explicit base case at all – everything can be done in the inductive step. How can this work without a base case? The answer is that the base case is implicitly included in the inductive step. Consider the case n = 3: the inductive-step instance is M (1) ∧ M (2) ∧ M (3) → M (4).