Equivalence
In Chapter 5, we saw DeMorgan’s laws for quantifiers – that if we have some domain, and if P(x)
is a predicate for individuals in that domain, then the following statements are equivalent:
¬(∃ x P(x))
is equivalent to∀ x ¬P(x)
¬(∀ x P(x))
is equivalent to∃ x ¬P(x)
The process of proving that two predicate logic statements are equivalent is the same as it was in propositional logic – we must prove the second proposition using the first as a premise, and we must prove the first given the second as a premise.
Example - how to prove equivalence
For example, to prove that ¬(∃ x P(x))
is equivalent to ∀ x ¬P(x)
, we must prove the sequents:
(
¬(∃((x: T) => P(x)))
)
⊢
(
∀((x: T) => ¬P(x))
)
and
(
∀((x: T) => ¬P(x))
)
⊢
(
¬(∃((x: T) => P(x)))
)
We prove both directions below:
(
¬(∃((x: T) => P(x)))
)
⊢
(
∀((x: T) => ¬P(x))
)
Proof(
1 ( ¬(∃((x: T) => P(x))) ) by Premise,
2 Let ( (a: T) => SubProof(
3 SubProof(
4 Assume ( P(a) ),
5 ( ∃((x: T) => P(x)) ) by ExistsI[T](3),
6 ( F ) by NegE(4, 1)
),
7 ( ¬P(a) ) by NegI(3)
)),
8 ( ∀((x: T) => ¬P(x)) ) by AllI[T](2)
)
And:
(
∀((x: T) => ¬P(x))
)
⊢
(
¬(∃((x: T) => P(x)))
)
Proof(
1 ( ∀((x: T) => ¬P(x)) ) by Premise,
2 SubProof(
3 Assume ( ∃((x: T) => P(x)) ),
4 Let ( (a: T) => SubProof(
5 Assume ( P(a) ),
6 ( ¬P(a) ) by AllE[T](1),
7 ( F ) by NegE(5, 6),
)),
8 ( F ) By ExistsE[T](3, 4)
),
9 ( ¬(∃((x: T) => P(x))) ) by NegI(2)
)
More extensive list of equivalences
Here is a more extensive list of equivalences in predicate logic. The remaining proofs are left as exercises for the reader:
¬(∃ x P(x))
is equivalent to∀ x ¬P(x)
¬(∀ x P(x))
is equivalent to∃ x ¬P(x)
∀ x (P(x) → ¬Q(x))
is equivalent to¬(∃ x P(x) ∧ Q(x))
∀ x ∀ y P(x, y)
is equivalent to∀ y ∀ x P(x, y)
∃ x ∃ y P(x, y)
is equivalent to∃ y ∃ x P(x, y)
Q ∧ (∀ x P(x))
is equivalent to∀ x (Q ∧ P(x))
(wherex
does not appear inQ
)Q v (∀ x P(x))
is equivalent to∀ x (Q V P(x))
(wherex
does not appear inQ
)Q ∧ (∃ x P(x))
is equivalent to∃ x (Q ∧ P(x))
(wherex
does not appear inQ
)Q V (∃ x P(x))
is equivalent to∃ x (Q V P(x))
(wherex
does not appear inQ
)