Lecture 13:  Some simple proofs I did in the class.

Def app:
(app x y)
= 
(if (endp x)
    y
    (cons (car x)
          (app (cdr x) y)))


Question 189:
(app a b) = (app b a)

This is not a theorem.  A counterexample is a = 12 and b = nil.
(app a b) = (app 12 nil) = nil
(app b a) = (app nil 12) = 12
---
Proof:  I'll simplify the lhs and then the rhs.

lhs
= 
(app a b)
= {def app}
(if (endp a)
    b
    (cons (car a)
          (app (cdr a) b)))

rhs
=
(if (endp b)
    a
    (cons (car b)
          (app (cdr b) a)))

Case 1 (endp a) <--> t
lhs = b
rhs = (if (endp b)
          a
        (cons (car b)
              (app (cdr b) a)))

 Case 1.1 (endp b) <--> t
  lhs = b
  rhs = a

???  Ah ha!  A counterexample is one where a and b are both
endp and different.  So let a = 12 and b = nil.

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Lemma endp-nil:  (endp nil) = t

Proof:
(endp nil)
= {def endp}
(not (consp nil)
= {comp, twice}
t
Q.E.D.

Lemma if-t: (if t x y) = x

Question 190:
(app nil b) = b

Proof:
(app nil b)
=  {def app}
(if (endp nil)
    b
    (cons (car nil)
          (app (cdr nil) b)))
= {endp-nil}
(if t
    b
    (cons (car nil)
          (app (cdr nil) b)))
= {if-t}
b
Q.E.D.
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Question 199:
(R & Q) --> (Q --> R)

Proof:
(R & Q) --> (Q --> R)
<--> {hyp R}
(R & Q) --> (Q --> T)
<--> {short circuit}
(R & Q) --> T
<--> {short circuit}
T
Q.E.D.

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Question 200:
(A --> B) v (B --> A)

Proof:
(A --> B) v (B --> A)
<--> {implicative disjunction, twice}
(!A v B) v (!B v A)
<--> {assoc}
!A v (B v (!B v A))
<--> {commut, twice}
!A v ((A v !B) v  B)
<-->
(!A v A) v (!B v B)
<--> {basic}
T v (!B v B)
<--> {short circuit}
T 
Q.E.D.

New Theorem:
(A --> B) v (C --> A)

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Lemma p-and-not-p
(P & !P) <--> NIL

Proof:
(P & !P)
<--> {DeMorgan, Double Negation}
!(!P v P)
<--> {basic}
!T {comp}
<-->
NIL
Q.E.D.

Question 201:
(R --> !P) --> ((P & R) --> S)

Proof:
(R --> !P) --> ((P & R) --> S)
<--> {Promotion}
(R --> !P) & (P & R) --> S
<--> {assoc}
(R --> !P) & P & R --> S
<--> {Hyp R}
(T --> !P) & P & R --> S
<--> {short circuit}
(!P & P & R) --> S
<--> {p-and-not-p}
(NIL & R) --> S
<--> {short circuit}
NIL --> S
<--> {short circuit}
T
Q.E.D.
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Question:
(P a) & (a = b) --> (P b)

Proof:
(P a) & (a = b) --> (P b)
<--> {Hyp a=b}
(P a) & (a = b) --> (P a)
<-->  {basic: x & y --> x}
T
Q.E.D.
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Question:
(P a) & (a = b) --> (Q b)

Proof:
(P a) & (a = b) --> (Q b)
<--> {hyp a=b}
(P a) & (a = b) --> (Q a)
???

Ah ha!  If I don't know anything else about P and Q, then this
isn't a theorem.  For example, suppose
(defun P (x) t)
(defun Q (x) nil)
a = 1 and b = 1.

Thus, 
(P a) & (a = b) --> (Q b)
<-->
(P 1) & (1 = 1) --> (Q 1)
<-->
(P 1) --> (Q 1)
<--> {comp}
T --> NIL
<--> {comp}
NIL
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Question:   Suppose it is a theorem that

T1:  (f x) = x

Is the following a theorem?

(P a) --> (P (f a))

Proof:
(P a) --> (P (f a))
<--> {Rewrite T1}
(P a) --> (P a)
<--> {basic}
T
Q.E.D.


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Question:   Suppose it is a theorem that

T1: (P x) --> (f x) = x

let sigma={x <-- a}, relieve the hyp (P a), check the equivalence

Is the following a theorem?

(P a) --> (P (f a))

Proof:
(P a) --> (P (f a))
<--> {Rewrite T1}
(P a) --> (P a)
...
[the rest is obvious given the proof above.]
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