page080_example8

hughmcguire

Proof of the following formula: 

if n² is odd,  then  n is odd

(#) Comments Suppositions and derivations Theorem and subgoals
    We presuppose the following:  
(1) DEFINITION1 (parenthesized) 
∀x[x is even iff not (x is odd)]
 
(2) DEFINITION 1 
∀x[x is even  means  ∃y(x = 2y)]
 
       
      We start working with the formula being proved
as follows:
(3) page080_example8   
if n² is odd,  then  n is odd
          ≡      by rewriting the form "if φ1 then φ2"
                to "if not φ2 then not φ1"
(4)     
if not (n is odd) then not (n² is odd)
    We'll assume that formula's antecedent:  
(5)   
not (n is odd)
 
      and we'll work on proving the consequent:
(6)     
not (n² is odd)
          ≡      by (1) with "x := n²"
(7)     
n² is even
          ≡      by (2) with "x := n²"
(8)     
∃y(n² = 2y)
      Removing that variable quantification
clarifies that to prove that formula,
it will suffice to find a value for the variable.
(9)     
n² = 2y
    Applying formula (1) to formula (5)
with "x := n"
yields: 		 
(10)   
n is even
 
        ≡      by (2) with "x := n"  
(11)   
∃y(n = 2y)
 
    That formula indicates there exists
some value say "k" satisfying that
quantified formula, i.e.: 		 
(12)   
n = 2k
 
       
      We'll work on transforming the left-hand side of formula (9)
to the right-hand side as follows:
(13)     
n²
          =      by (12)
(14)     
(2k)²
          =      by definition of "τ²"
(15)     
(2k)(2k)
          =      by simplifying
(16)     
4k²
          =      by basic algebra/arithmetic
(17)     
2(2k²)
      And that satisfies our earlier goal (9)
with "y := 2k²"; i.e. we have:
(18)     
true
      That concludes this part of the proof.
       
      Thus, the theorem that was given is true.

identification:
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1233837259913
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