I'm practicing writing FOL sentences and I have the following sentences. The first sentence is the solution I provided and the second is the correct solution ( per the solutions manual).The difference is where I've placed ∃z. I understand there is a difference in scope but I don't understand how that difference is applied to the predicates.

  1. ∀x (Txi => ∃y∃z(Txy ∧ Tyz ∧ Vz))
  2. ∀x (Txi => ∃y(Txy ∧ ∃z(Tyz ∧ Vz))

1 Answer 1


Both solutions are correct: Since z does not occur free in Txy, the ∃z has no effect on it, and the two formalizations are logically equivalent.

In general, you can move existential and universal quantifiers between the right-hand side of a conjunction resp. implication and the front, provided that the left-hand side does not contain free occurrences of the binding variable:

A ∧ ∃xB ≡ ∃x(A ∧ B), if x not free in A
A → ∀xB ≡ ∀x(A → B), if x not free in A

You can find a more comprehensive list of when formulas with the quantifier outside vs. inside a subformula are logically equivalent e.g. on the Wikipedia article on prenex normal form.

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