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
• 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).