page077_Example1

mcguireh

Proof of the following formula: 

If n is odd,  then  n² is odd

(#) Comments Suppositions and derivations Theorem and subgoals
    We presuppose the following:  
(1) DEFINITION 1 
∀x[x is odd  means  ∃y(x = 2y + 1)]
 
(2)   
∀x[x² = x·x]
 
       
      We start working with the formula being proved
as follows:
(3) page077_Example1   
If n is odd,  then  n² is odd
    We'll assume that formula's antecedent:  
(4)   
n is odd
 
      and we'll work on proving the consequent:
(5)     
n² is odd
          ≡      by (1) with "x := n²"
(6)     
∃y(n² = 2y + 1)
      Removing that variable quantification
clarifies that to prove that formula,
it will suffice to find a value for the variable.
(7)     
n² = 2y + 1
    Applying the equivalence in formula (1)
to formula (4)
with "x := n"
yields: 		 
(8)   
∃y(n = 2y + 1)
 
    That formula indicates there exists
some value say "k" satisfying that
quantified formula, i.e.: 		 
(9)   
n = 2k + 1
 
       
      We'll work on transforming the left-hand side of formula (7)
to the right-hand side as follows:
(10)     
n²
          =      by (9)
(11)     
(2k+1)²
          =      by binomial expansion
(12)     
(2k)² + 4k + 1
          =      by (2) with "x := 2k"
(13)     
[(2k)·(2k)] + 4k + 1
          =      by simplifying
(14)     
[4·k²] + 4k + 1
          =      by basic algebra/arithmetic
(15)     
2(2k² + 2k) + 1
      And that satisfies our earlier goal (7)
with "y := 2k² + 2k"; i.e. we have:
(16)     
true
      That concludes this part of the proof.
       
      Thus, the theorem that was given is true.

identification:
1233598939718
1233598214531
725187