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).