page081_example11

hughmcguire

Proof of the following formula: 

if 3n + 2 is odd,  then  n is odd

(#) Comments Suppositions and derivations Theorem and subgoals
    We presuppose the following:  
(1) "Definition1" (parenthesized) 
∀x[x is even  if and only if  not (x is odd)]
 
(2) "Definition1" 
∀x[x is even  means  ∃y(x = 2y)]
 
       
      We start working with the formula being proved
as follows:
(3) page081_example11   
if 3n + 2 is odd,  then  n is odd
    Attempting a proof by contradiction,
suppose the negation of that formula: 		 
(4)   
not (if 3n + 2 is odd,  then  n is odd)
 
        ≡      by rewriting the form "not (if φ1 then φ2)"
                to "(φ1 and not φ2)" 		 
(5)   
3n + 2 is odd  and  not (n is odd)
 
    That formula can be separated into two
alternative suppositions/derivations; first: 		 
(6)   
3n + 2 is odd
 
    and also:  
(7)   
not (n is odd)
 
        ≡      by (1) "Definition1" with "x := n"  
(8)   
n is even
 
        ≡      by (2) "Definition1" with "x := n"  
(9)   
∃y(n = 2y)
 
    That formula indicates there exists
some value say "k" satisfying that
quantified formula, i.e.: 		 
(10)   
n = 2k
 
        ≡      by basic algebra/arithmetic  
(11)   
3n + 2 = 3(2k) + 2
 
        ≡      by basic algebra/arithmetic  
(12)   
3n + 2 = 2(3k + 1)
 
    Since "3k + 1" satisfies that formula,
we can say some value satisfies it, i.e.: 		 
(13)   
(∃t)[3n + 2 = 2t]
 
        ≡      by (2) "Definition1" with "y := t, x := 3n+2, y := t"  
(14)   
(3n + 2) is even
 
        ≡      by (1) "Definition1" with "x := 3n+2"  
(15)   
not ((3n + 2) is odd)
 
        ↓      by (6)  
(16)   
not true
 
        ≡      by simplifying  
(17)   
false
 
    Thus, the earlier supposition of
the negation of the goal to prove
yields a contradiction (false).  Therefore,
the original theorem that was given
must be true. 		 

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