2.1. First-Order Logic 11 have ϕ without ψ). Clearly ϕ → ψ should be true when both formulas are true, and it should be false if ϕ is true but ψ is false. It is maybe not so clear what to do when ϕ is false this is clarified by rephrasing implication as disjunction (which is often how it is defined in the first place). ϕ → ψ means either ψ holds or ϕ fails i.e., ψ ∨ (¬ϕ). The truth of that statement lines up with our assertions earlier, and gives truth values for when ϕ is false – namely, that the implication is true. Another way to look at this is to say ϕ → ψ is only false when proven false i.e., when it has a true antecedent but a false consequent. From this it is clear that ¬(ϕ → ψ) is ϕ & (¬ψ). There is an enormous difference between implication in natural language and implication in logic. Implication in natural language tends to connote causation, whereas the truth of ϕ → ψ need not give any connection at all between the meanings of ϕ and ψ. It could be that ϕ is a contradiction, or that ψ is a tautology. Also, in natural language we tend to dismiss implications as irrelevant or meaningless when the antecedent is false, whereas to have a full and consistent logical theory we cannot throw those cases out. Example 2.1.2. The following are true implications: • If fish live in the water, then earthworms live in the soil. • If rabbits are aquamarine blue, then earthworms live in the soil. • If rabbits are aquamarine blue, then birds drive cars. The negation of the final statement is “Rabbits are aquamarine blue but birds do not drive cars.” The statement “If fish live in the water, then birds drive cars” is an example of a false implication. Equivalence is two-way implication and indicated by a double- headed arrow: ϕ ↔ ψ or ϕ ⇔ ψ. It is an abbreviation for (ϕ → ψ) & (ψ → ϕ), and is true when ϕ and ψ are either both true or both false. Verbally we might say “ϕ if and only if ψ”, which is often abbreviated to “ϕ iff ψ”. In terms of just conjunction, disjunction, and negation, we may write equivalence as (ϕ & ψ) ∨ ((¬ϕ) & (¬ψ)). Its negation is exclusive or, (ϕ ∨ ψ) & ¬(ϕ & ψ).

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