intro year 2- don't be caught out!

Question 1

∀I?

Answer 1

t | | Φt (assumption would be line under)or reiteration
| | |
| | |…
| | |
| |ψt
|(∀x) ψx
immediately to the right of the given scope line there is a t barrier which is not introduced as a scope line of an assumption
Φt exists to the right of the t barrier
this is proven from starting with assumption or reiteration
and the variable must be t
this must then be changed to (∀x)Φx - these must have the same variable
the line of Φt is the only thing that justifies (∀x)Φx with
∀I

Question 2

∃E?

Answer 2

(∃x)Φx
t | Φt (an immediate assumption) A
|
|
| ψy
ψy
(∃x)Φx is already on the same scope line
immediately to the right of the scope bar there is a subproof from Φt to ψy when t is any parameter not occuring in (∃x)Φx
when at ψy, the we can move ψy out of the subproof when not containing parameter t.
parameter t cannot leave the scope
appeal to line of (∃x)Φx then start of assumption to end of assumption ∃E

Question 3

∀E?

Answer 3

(∀x) Φx
Φa 
when  
(∀x) Φx is already on the same scope line you can remove (∀x)  for Φa as long as every variable is replaced by the same term 
appeal to line of (∀x) Φx then ∀E

Question 4

∃I?

Answer 4

Ha
(∃x)Hx

Ha ^ Fa
(∃x)Hx ^ fa
or
(∃x)Ha^ Fa

a formula which contains a term must be on the same scope line
then you can add (∃x) to it if you replace one or more occurances of the term with x- the original variable
justified with appeal to original formula and then ∃I