In today's class I proved several theorems.  Here I only
recount one of them.  The definitions are

(defun mem (e x)
  (if (endp x)
      nil
    (if (equal e (car x))
        t
      (mem e (cdr x)))))

(defun nth (i x)
  (if (zp i)
      (car x)
    (nth (- i 1) (cdr x))))

Theorem:
(mem e x) --> (exists i : (natp i) & (< i (len x)) & (nth i x)e)

Proof:  Use Exists-concl and replace i by (index e x), where:

(defun index (e x)
  (if (endp x)
      0
    (if (equal e (car x))
        0
      (+ 1 (index e (cdr x))))))

Thus, our goal is

(mem e x) -->  (natp (index e x)) & (< (index e x) (len x)) & (nth (index e x) x)e

We clearly need to prove three things:

(a) (mem e x) -->  (natp (index e x)) & (< (index e x) (len x)) & (nth (index e x) x)e)
(b) (mem e x) -->  (< (index e x) (len x))
(c) (mem e x) -->  (nth (index e x) x)e

I will prove just (c).

(mem e x) -->  (nth (index e x) x)e

Induct on x.

Base Case:  (endp x) --> (mem e x) --> (nth (index e x) x)e
which is trivial because (mem e x) is false.

Induction Step:
 ~(endp x)
 &
 ((mem e (cdr x)) -->  (nth (index e (cdr x)) (cdr x))e)
-->
 (mem e x) -->  (nth (index e x) x)e

By promotion we get
 ~(endp x)
 &
 ((mem e (cdr x)) -->  (nth (index e (cdr x)) (cdr x))e)
 &
 (mem e x)
--> 
 (nth (index e x) x)e

Expanding (mem e x):

 ~(endp x)
 &
 ((mem e (cdr x)) -->  (nth (index e (cdr x)) (cdr x))e)
 &
 (e(car x) | (mem e (cdr x)))
--> 
 (nth (index e x) x)e

We consider two cases:

Case 1:  e(car x).  The formula we are proving is

 ~(endp x)
 &
 ((mem e (cdr x)) -->  (nth (index e (cdr x)) (cdr x))e)
 &
 e(car x)
--> 
 (nth (index e x) x)e

But expanding (index e x) gives 0 hyps 1 and 3; and (nth 0 x) is (car x), which
is e by the last hyp.

Case 2: e!(car x) & (mem e (cdr x)):

The formula we are proving is

 ~(endp x)
 &
 ((mem e (cdr x)) -->  (nth (index e (cdr x)) (cdr x))e)
 &
 e!(car x)
 &
 (mem e (cdr x))
--> 
 (nth (index e x) x)e

But now we can forward chain from hyp 4 into hyp 2 to get:

 ~(endp x)
 &
 (nth (index e (cdr x)) (cdr x))e
 &
 e!(car x)
 &
 (mem e (cdr x))
--> 
 (nth (index e x) x)e

Then, expanding (index e x) under hyps 1 and 3, and then expanding the
containing nth and using arithmetic we get

 ~(endp x)
 &
 (nth (index e (cdr x)) (cdr x))e
 &
 e!(car x)
 &
 (mem e (cdr x))
--> 
 (nth (index e (cdr x)) (cdr x))e

Which is a tautology.

Q.E.D.