∀x ∀y (P x → (P y → ¬x = y)) ⊢ ¬∃x P x
I will prove ∀x ∀y (Px → (P y → ¬x = y)) ⊢ ¬∃x Px.
∀x ∀y (P x → (P y → ¬x = y)) ⊢ ¬∃x P x
I will prove ∀x ∀y (Px → (P y → ¬x = y)) ⊢ ¬∃x Px.
By instantiating the variables x and y in the premiss with the
same constant, one can show that any object satisfying P must be
different from itself. Hence there can’t be such an object.
∀x ∀y (P x → (P y → ¬x = y)) ⊢ ¬∃x P x
I will prove ∀x ∀y (Px → (P y → ¬x = y)) ⊢ ¬∃x Px.
By instantiating the variables x and y in the premiss with the
same constant, one can show that any object satisfying P must be
different from itself. Hence there can’t be such an object.
For the Natural Deduction proof I assume Pa (The plan is to
apply ∃Elim later using the assumption ∃x Px). Then I derive
from Pa and the premiss a contradiction. Then I assume ∃x Px
and discharge Pa by applying ∃Elim. By applying ¬Intro I
conclude ¬∃x Px.
∀x ∀y (P x → (P y → ¬x = y)) ⊢ ¬∃x P x
∀x ∀y (Px → (P y → ¬x = y))
This is the premiss.
∀x ∀y (P x → (P y → ¬x = y)) ⊢ ¬∃x P x
∀x ∀y (Px → (P y → ¬x = y))
∀y (Pa → (P y → ¬a = y))
Pa → (Pa → ¬a = a)
I apply ∀Elim twice using the same constant.
∀x ∀y (P x → (P y → ¬x = y)) ⊢ ¬∃x P x
Pa
∀x ∀y (Px → (P y → ¬x = y))
∀y (Pa → (P y → ¬a = y))
Pa → (Pa → ¬a = a)
Later I will assume ∃x Px and show that this leads
to a contradiction. So I assume Pa. . .
∀x ∀y (P x → (P y → ¬x = y)) ⊢ ¬∃x P x
Pa
and apply →Elim.
∀x ∀y (Px → (P y → ¬x = y))
∀y (Pa → (P y → ¬a = y))
Pa → (Pa → ¬a = a)
Pa → ¬a = a
∀x ∀y (P x → (P y → ¬x = y)) ⊢ ¬∃x P x
Pa
Pa
This is repeated,
∀x ∀y (Px → (P y → ¬x = y))
∀y (Pa → (P y → ¬a = y))
Pa → (Pa → ¬a = a)
Pa → ¬a = a
∀x ∀y (P x → (P y → ¬x = y)) ⊢ ¬∃x P x
Pa
which gives me ¬a = a
∀x ∀y (Px → (P y → ¬x = y))
∀y (Pa → (P y → ¬a = y))
Pa
Pa → (Pa → ¬a = a)
Pa → ¬a = a
¬a = a
∀x ∀y (P x → (P y → ¬x = y)) ⊢ ¬∃x P x
Pa
[a = a]
This is an application of =Intro.
∀x ∀y (Px → (P y → ¬x = y))
∀y (Pa → (P y → ¬a = y))
Pa
Pa → (Pa → ¬a = a)
Pa → ¬a = a
¬a = a
∀x ∀y (P x → (P y → ¬x = y)) ⊢ ¬∃x P x
Pa
[a = a]
∀x ∀y (Px → (P y → ¬x = y))
∀y (Pa → (P y → ¬a = y))
Pa
Pa → (Pa → ¬a = a)
Pa → ¬a = a
¬a = a
¬∃x Px
I apply ¬Intro to ∃x Px. I can’t discharge any
assumption because I didn’t assume ∃x Px.
∀x ∀y (P x → (P y → ¬x = y)) ⊢ ¬∃x P x
Pa
[a = a]
∃x Px
¬∃x Px
Following my plan, I assume ∃x Px.
∀x ∀y (Px → (P y → ¬x = y))
∀y (Pa → (P y → ¬a = y))
Pa
Pa → (Pa → ¬a = a)
Pa → ¬a = a
¬a = a
∀x ∀y (P x → (P y → ¬x = y)) ⊢ ¬∃x P x
[Pa]
[a = a]
∃x Px
∀x ∀y (Px → (P y → ¬x = y))
∀y (Pa → (P y → ¬a = y))
[Pa]
Pa → (Pa → ¬a = a)
Pa → ¬a = a
¬a = a
¬∃x Px
¬∃x Px
This is a correct application of ∃Elim because the
only undischarged assumption containing a is Pa
and ¬∃x Px doesn’t contain a. So I can discharge
the assumptions of Pa.
∀x ∀y (P x → (P y → ¬x = y)) ⊢ ¬∃x P x
[Pa]
[a = a]
∃x Px
∀x ∀y (Px → (P y → ¬x = y))
∀y (Pa → (P y → ¬a = y))
[Pa]
Pa → (Pa → ¬a = a)
Pa → ¬a = a
¬a = a
¬∃x Px
¬∃x Px
I still need to discharge the assumption ∃x Px. To
this end I assume it and obtain a contradiction
∃x Px
∀x ∀y (P x → (P y → ¬x = y)) ⊢ ¬∃x P x
∀x ∀y (Px → (P y → ¬x = y))
∀y (Pa → (P y → ¬a = y))
[Pa]
Pa → (Pa → ¬a = a)
Pa → ¬a = a
¬a = a
[Pa]
[a = a]
[∃x Px]
¬∃x Px
¬∃x Px
[∃x Px]
¬∃x Px
Using ¬Intro I discharge both assumptions of
∃x Px and the proof is complete..