Understanding First-order Logic 139 Proof Again, we first formalize the argument in first-order logic by introducing predicates: B (x) meaning “x is black” and W (x) meaning “x is white.” Now the argument can be formalized as follows: ∀x(B (x) ∨ W (x)) |= ∀xB (x) ∨ ∀xW (x) To falsify this logical consequence it is sufficient to find any counter-model, that is, a structure S and a variable assignment v such that S , v |= ∀x(B (x) ∨ W (x)) and S , v ∀xB (x) ∨ ∀xW (x) Note that the interpretations of the predicates B and W need not have anything to with black, white, or any colors in that structure We therefore choose to take the structure S with domain the set Z of all integers, where the interpretation of B is the predicate Even, where Even(n) means “n is even”, and the interpretation of W is the predicate Odd, where Odd(n) means “n is odd.” We already know that the variable assignment is irrelevant here because there are no free variables in this argument, but we will nevertheless need it in order to process the truth definitions of the quantifiers Consider any variable assignment v in S Then take any variable assignment v that differs from v possibly only on x The integer v (x) is even or odd, therefore S , v |= B (x) or S , v |= W (x), hence (by the truth definition of ∨): S , v |= B (x) ∨ W (x) Thus, by the truth definition of ∀, it follows that: S , v |= ∀x(B (x) ∨ W (x)) On the other hand, consider the variable assignment v1 obtained from v by redefining it on x as follows: v1 (x) = Then: S , v1 B (x) By the truth definition of ∀, it follows that: S , v ∀xB (x) Likewise, consider the variable assignment v2 obtained from v by redefining it on x as follows: v2 (x) = Then: (3) S , v2 W (x) By the truth definition of ∀, it follows that: (4) S , v ∀xW (x) By the truth definition of ∨, it follows that (5) S , v ∀xB (x) ∨ ∀xW (x) Therefore, by the truth definition of →, it follows from (2) and (5) that (6) S , v ∀x(B (x) ∨ W (x)) → (∀xB (x) ∨ ∀xW (x))