Existential Instantiation permits you to remove an existential quantifier from a formula which has an existential quantifier as its main connective. It is one of those rules which involves the adoption and dropping of an extra assumption (like ∼I,⊃I,∨E, and ≡I).
The circumstance that Existential Instantiation gets invoked looks like this.
(∃x)<scope> ? ? ? << <target> ?
That is, a formula you have has an existential quantifier as its main connective-- and you are trying to obtain some formula which we will call <target>.
For example,
(∃x)(Fx) ? ? ? << (∃y)(Fy) ?
invites the use of EI. In this case (∃x)(Fx) is the existential formula, (Fx) is its scope, and (∃y)(Fy) is the target.
To use EI you follow a three stage process. First you assume the <scope>. The illustrated fragment will then become
(∃x)<scope> ?
<scope> Ass ? ? << <target> ?
Second you continue deriving in this new context until you manage to derive <target> thus
(∃x)<scope> ?
<scope> Ass
<proof fragment in here > ?
<target> ? ? ? << <target> ?
Finally you select <target> at the end of the sub-proof, select the existential formula, and click EI to complete the existential instantiation.
n (∃x)<scope> ?
<scope> Ass
<proof fragment in here > ? k
<target> ? <target> n,k EI
In the example, the derivation would look like
1 (∃x)Fx Ass 2
Fx Ass 3
(∃y)Fy 2 EG 4 (∃y)Fy 1,3 EI
There are restrictions on EI. The variable of quantification must not be free either
More about these later.
Proofs
It is probably easier to learn EI using Tactics. After you start each derivation,switch Tactics on, select the existentially quantified formula, and click EI.
Derive the following
a) (∃x)(Fa) ∴ Fa
b) (∃x) (Fx∧Gx) ∴ (∃x)(Fx )
c) (∃x) (Fx∧Gx) ∴ (∃y)(Gy)
d) (∃x)(Fx)∴ (∃y)(Fy)
e) (∃x)(Fx∧Gx) ∴ (∃y)(Fy)∧(∃z)(Gz)
Proofs
Repeat the derivations of Exercise 1 without using Tactics.
a) (∃x)(Fa) ∴ Fa
b) (∃x) (Fx∧Gx) ∴ (∃x)(Fx )
c) (∃x) (Fx∧Gx) ∴ (∃y)(Gy)
d) (∃x)(Fx)∴ (∃y)(Fy)
e) (∃x)(Fx∧Gx) ∴ (∃y)(Fy)∧(∃z)(Gz)
Proofs
Each of following derivations will use Existential Instantiation at some stage.
Derive
a) ∴ (∃x)Fx≡(∃y)Fy
b) (∀x)Fx,(∃x)~Fx∴G
c) ∴ (∃x) (Fx∨Gx) ≡ (∃y)Fy∨(∃z)Gz
d) (∃x)(Fx∨Gx),(∀x)(Fx⊃Hx),(∀x)(Gx⊃Hx)∴(∃x)Hx
e) ~(∃x)(Fx∧~Gx),~(∃x)(Gx∧~Hx)∴(∀x)(Fx⊃Hx)
f) ~(∃x)Gx,(∀x)(Fx⊃Gx)∴(∀x)(Fx⊃Hx)
g) (∃x)Fx∨(∃x)Gx,(∀x)((Fx∨Gx)⊃Hx)∴(∃x)Hx