In this proof, I use ``alpha < beta'' to denote the idea that
alpha can be proved by proving beta.  Unexplained steps are just
the usual propositional calculus.

In the proof below I do an abbreviated kind of case analysis.
If you are proving

(a | b) --> c

you can prove it by proving

Case 1:
a --> c

Case 2:
b --> c

You should be able to justify this step by our usual rules and tautologies.

-----------------------------------------------------------------

Theorem:
(sub x y) --> (all e : (mem e x) --> (mem e y))

Proof.  By induction on x.

BC: (endp x)

(endp x) --> (sub x y) --> (all e : (mem e x) --> (mem e y))
<-->
(endp x) & (sub x y) --> (all e : (mem e x) --> (mem e y))
<--> {def sub}
(endp x) --> (all e : (mem e x) --> (mem e y))
> {all-concl}
(endp x) -->  (mem e x) --> (mem e y)
<-->
(endp x) & (mem e x) --> (mem e y)
<--> {def mem}
(endp x) & nil --> (mem e y)
<-->
t

IS: !(endp x)
    &
    [(sub (cdr x) y) --> (all e : (mem e (cdr x)) --> (mem e y))]    
  -->
    (sub x y) --> (all e : (mem e x) --> (mem e y))

<-->

 !(endp x)
 &
 [(sub (cdr x) y) --> (all e : (mem e (cdr x)) --> (mem e y))]    
 &
 (sub x y)
--> 
 (all e : (mem e x) --> (mem e y))

<--> {def sub}

 !(endp x)
 &
 [(sub (cdr x) y) --> (all e : (mem e (cdr x)) --> (mem e y))]    
 &
 (mem (car x) y)
 &
 (sub (cdr x) y)
--> 
 (all e : (mem e x) --> (mem e y))

<-->

 !(endp x)
 &
 (all e : (mem e (cdr x)) --> (mem e y))    
 &
 (mem (car x) y)
 &
 (sub (cdr x) y)
--> 
 (all e : (mem e x) --> (mem e y))

<--> {all-concl}

 !(endp x)
 &
 (all e : (mem e (cdr x)) --> (mem e y))    
 &
 (mem (car x) y)
 &
 (sub (cdr x) y)
--> 
 (mem e x) --> (mem e y)

<--> 

 !(endp x)
 &
 (all e : (mem e (cdr x)) --> (mem e y))    
 &
 (mem (car x) y)
 &
 (sub (cdr x) y)
 &
 (mem e x)
--> 
 (mem e y)

<--> {def mem}

 !(endp x)
 &
 (all e : (mem e (cdr x)) --> (mem e y))    
 &
 (mem (car x) y)
 &
 (sub (cdr x) y)
 &
 [e(car x)  | (mem e (cdr x))
--> 
 (mem e y)

< {Cases}

Case 1: 
 e(car x)
 & 
 !(endp x)
 &
 (all e : (mem e (cdr x)) --> (mem e y))    
 &
 (mem (car x) y)
 &
 (sub (cdr x) y)
 &
 [e(car x)  | (mem e (cdr x))
--> 
 (mem e y)

<--> {rewrite concl with hyp 1}

T

Case 2: 
 (mem e (cdr x))
 & 
 !(endp x)
 &
 (all e : (mem e (cdr x)) --> (mem e y))    
 &
 (mem (car x) y)
 &
 (sub (cdr x) y)
 &
 [e(car x)  | (mem e (cdr x))
--> 
 (mem e y)

< {all-hyp}

 (mem e (cdr x))
 & 
 !(endp x)
 &
 (mem e (cdr x)) --> (mem e y)    
 &
 (mem (car x) y)
 &
 (sub (cdr x) y)
 &
 [e(car x)  | (mem e (cdr x))
--> 
 (mem e y)

<-->

 (mem e (cdr x))
 & 
 !(endp x)
 &
 (mem e y)    
 &
 (mem (car x) y)
 &
 (sub (cdr x) y)
 &
 [e(car x)  | (mem e (cdr x))
--> 
 (mem e y)

<-->

T

Q.E.D.