Consider the sentence "If all the birds in my house are peacocks, then Charlie is a peacock," and let the provided symbols be denoted as B for bird, H for house, P for peacock, and c for Charlie.

Are the following two translations exactly equivalent?

  1. (x)( ( (Bx & Hx) → Px) → Pc)
  2. ((x)( ( (Bx & Hx) → Px)) → Pc

If they are different, what is the exact difference? In particular, what is an example that works for (1) and not (2) (and vice versa). If they are equivalent, how can I generalize the situations for which the correctness of the translation is not changed by moving the position of the parentheses?

As a note, the textbook that I am using denotes (x) as a universal qualifier instead of using (∀x).

  • 1
    They are not equivalent. To move quantifiers out of implications correctly see prenex normal form: (∀x ϕ) → ψ is equivalent to ∃x(ϕ → ψ), not to ∀x(ϕ → ψ).
    – Conifold
    Nov 13 '19 at 18:53
  • a priori that Charlie is a bird. Nov 14 '19 at 4:49
  • Well 2 isn't a well-formed formula. Did you mean to add another closed parenthesis? Moreover there is an effective procedure for adding parenthesis such that these questions shouldn't arise. Nov 15 '19 at 19:18

They are not equivalent. Here is an example. Suppose one (but not all) of the birds in your house is a peacock, and that Charlie isn't a peacock. Then:

  1. (Bx & Hx) → Px) = T for some x, and Pc = F.
  2. So, (x)(((Bx & Hx) → Px) → Pc) = F.
  3. But ((x)(((Bx & Hx) → Px)) = F.
  4. Therefore, ((x)(((Bx & Hx) → Px)) → Pc = T.

2 and 4 show that the translations are not equivalent. In general, (x)(Ax → B) is not equivalent to (x)(Ax) → B. (You can see more on this here.) In your case, the latter formulation is correct. Your statement is conditional: if [...], then [...]. So its main connective should be →, and it should connect "all the birds are ..." and "Charlie is ...". So the correct form is (x)(Ax) → B.

Another way to see this is to note that (x)(Ax → B) says Ax → B about each individual x. That is, it is like saying: (Aa → B) & (Ab → B) & (Ac → B) & .... But this is not what you want to say -- you don't want to say that if that bird is a peacock then Charlie is a peacock, and if this bird is a peacock then Charlie is a peacock, and so on for every single bird. You want to say that if everyone is then Charlie is.

A good rule of thumb to get the correct translations in these kinds of cases is to have your quantifier range only over where its variable occurs. For example, in (x)(((Bx & Hx) → Px) → Pc), the variable x doesn't occur in 'Pc', so according to this rule of thumb the (x) quantifier shouldn't range over it.

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