138 Logic as a Tool ∃xP (x) ∧ ∃xQ(x) ∃x(P (x) ∧ Q(x)) Indeed, the structure N obtained from N where P (x) is interpreted as ‘x is even’ and Q(x) is interpreted as ‘x is odd’ is a counter-model: ∃x(P (x) ∧ Q(x)) N |= ∃xP (x) ∧ ∃xQ(x), but N Example 111 Here we illustrate “semantic reasoning”, based on the formal semantics of first-order formulae, for proving or disproving first-order logical consequences Show the logical validity of the following argument: “If Tinkerbell is a Disney fairy and every Disney fairy has blue eyes, then someone has blue eyes.” Proof Let us first formalize the argument in first-order logic by introducing predicates: P (x) meaning “x is a Disney fairy”, Q(x) meaning “x has blue eyes”, and a constant symbol c interpreted as “Tinkerbell.” The argument can now be formalized as follows: P (c) ∧ ∀x(P (x) → Q(x)) |= ∃yQ(y ) To show its validity take any structure S for the first-order language introduced above and any variable assignment v in S Now, suppose S , v |= (P (c) ∧ ∀x(P (x) → Q(x)) We have to show that: S , v |= ∃yQ(y ) From the assumption (2) we have, by the truth definition of ∧, that: S , v |= P (c) and S , v |= ∀x(P (x) → Q(x)) Let v be a variable assignment obtained from v by redefining it on x as follows: v (x) = cS (Recall that cS is the interpretation of the constant symbol c in S.) Then: S , v |= P (x) According to the truth definition of ∀, it follows from (5) that: S , v |= P (x) → Q(x) From (6) and (7) it follows that: S , v |= Q(x) Now, let v be a variable assignment obtained from v by redefining it on y as follows: v (y ) = v (x) = cS We then have: S , v |= Q(y ) According to the truth definition of ∃, it follows from (9) that: (3) S , v |= ∃yQ(y ), so we are done Prove that the following argument is not logically valid: “If everything is black or white then everything is black or everything is white.”