¬∃x (P x ∧ ¬Q x) ⊢ ∀x (P x → Q x)
We want to show that ¬∃x (Px ∧ ¬Qx) ⊢ ∀x (Px → Qx).
¬∃x (P x ∧ ¬Q x) ⊢ ∀x (P x → Q x)
We want to show that ¬∃x (Px ∧ ¬Qx) ⊢ ∀x (Px → Qx).
I will prove the conclusion by proving Pa → Qa and applying
∀Intro.
¬∃x (P x ∧ ¬Q x) ⊢ ∀x (P x → Q x)
We want to show that ¬∃x (Px ∧ ¬Qx) ⊢ ∀x (Px → Qx).
I will prove the conclusion by proving Pa → Qa and applying
∀Intro.
To prove Pa → Qa I assume its negation ¬(Pa → Qa) and prove
from this Pa ∧ ¬Qa. Applying ∃Intro will give me
∃x (Px ∧ ¬Qx), which contradicts the premiss and I can
conclude Pa → Qa by applying ¬Elim.
¬∃x (P x ∧ ¬Q x) ⊢ ∀x (P x → Q x)
I write down the negation of
Pa → Qa and try to obtain Pa.
To this end I assume Pa and
¬Pa.
¬Pa
Pa
¬(Pa → Qa)
¬∃x (P x ∧ ¬Q x) ⊢ ∀x (P x → Q x)
Using ¬Elim I conclude Qa.
Pa
¬Pa
Qa
¬(Pa → Qa)
¬∃x (P x ∧ ¬Q x) ⊢ ∀x (P x → Q x)
Now I apply →Intro to
discharge Pa.
¬Pa
[Pa]
Qa
Pa → Qa
¬(Pa → Qa)
¬∃x (P x ∧ ¬Q x) ⊢ ∀x (P x → Q x)
This yields a contradiction,
which allows me to apply
¬Elim and ¬Pa is discharged
as well.
[¬Pa]
[Pa]
Qa
Pa → Qa
¬(Pa → Qa)
Pa
¬∃x (P x ∧ ¬Q x) ⊢ ∀x (P x → Q x)
Following my plan, I try to
prove ¬Qa from the
assumption ¬(Pa → Qa). I try
to do this by assuming Qa
hoping that I can apply ¬Intro.
[¬Pa]
[Pa]
Qa
Pa → Qa
Qa
¬(Pa → Qa)
Pa
¬(Pa → Qa)
¬∃x (P x ∧ ¬Q x) ⊢ ∀x (P x → Q x)
But that’s easy because I can
derive Pa → Qa by →Intro. . .
[¬Pa]
[Pa]
Qa
Pa → Qa
¬(Pa → Qa)
Pa
Qa
Pa → Qa
¬(Pa → Qa)
¬∃x (P x ∧ ¬Q x) ⊢ ∀x (P x → Q x)
and applying then ¬Intro.
[¬Pa]
[Pa]
Qa
Pa → Qa
¬(Pa → Qa)
Pa
[Qa]
Pa → Qa
¬(Pa → Qa)
¬Qa
¬∃x (P x ∧ ¬Q x) ⊢ ∀x (P x → Q x)
The plan is to prove
Pa ∧ ¬Qa. . .
[¬Pa]
[Pa]
Qa
Pa → Qa
[Qa]
¬(Pa → Qa)
Pa → Qa
Pa
Pa ∧ ¬Qa
¬(Pa → Qa)
¬Qa
¬∃x (P x ∧ ¬Q x) ⊢ ∀x (P x → Q x)
. . . and to apply ∃Intro,
[¬Pa]
[Pa]
Qa
Pa → Qa
[Qa]
¬(Pa → Qa)
Pa → Qa
Pa
Pa ∧ ¬Qa
∃x (Px ∧ ¬Qx)
¬(Pa → Qa)
¬Qa
¬∃x (P x ∧ ¬Q x) ⊢ ∀x (P x → Q x)
which contradicts the premiss.
[¬Pa]
[Pa]
Qa
Pa → Qa
[Qa]
¬(Pa → Qa)
Pa → Qa
¬(Pa → Qa)
Pa
¬Qa
Pa ∧ ¬Qa
∃x (Px ∧ ¬Qx)
¬∃x (Px ∧ ¬Qx)
¬∃x (P x ∧ ¬Q x) ⊢ ∀x (P x → Q x)
So using ¬Elim I infer
Pa → Qa and discharge
¬(Pa → Qa).
[¬Pa]
[Pa]
Qa
Pa → Qa
[Qa]
[¬(Pa → Qa)]
Pa → Qa
[¬(Pa → Qa)]
Pa
¬Qa
Pa ∧ ¬Qa
∃x (Px ∧ ¬Qx)
¬∃x (Px ∧ ¬Qx)
Pa → Qa
¬∃x (P x ∧ ¬Q x) ⊢ ∀x (P x → Q x)
Finally there is no
undischarged assumption left
that contains the constant a.
Thus, I can apply ∀Intro.
[¬Pa]
[Pa]
Qa
Pa → Qa
[Qa]
[¬(Pa → Qa)]
Pa → Qa
[¬(Pa → Qa)]
Pa
¬Qa
Pa ∧ ¬Qa
∃x (Px ∧ ¬Qx)
¬∃x (Px ∧ ¬Qx)
Pa → Qa
∀x (Px → Qx)
¬∃x (P x ∧ ¬Q x) ⊢ ∀x (P x → Q x)
Apart from the premiss there is
no undischarged assumption
left. So the proof is complete.
[¬Pa]
[Pa]
Qa
Pa → Qa
[Qa]
[¬(Pa → Qa)]
Pa → Qa
[¬(Pa → Qa)]
Pa
¬Qa
Pa ∧ ¬Qa
∃x (Px ∧ ¬Qx)
¬∃x (Px ∧ ¬Qx)
Pa → Qa
∀x (Px → Qx)