page077_Example2

mcguireh

Proof of the following formula: 

if (m is square) and (n is square),  then  mn is square

(#) Comments Suppositions and derivations Theorem and subgoals
    We presuppose the following:  
(1)   
∀a[a is square  means  ∃b(a = b²)]
 
(2)   
∀x[x² = x·x]
 
       
      We start working with the formula being proved
as follows:
(3) page077_Example2   
if (m is square) and (n is square),  then  mn is square
    We'll assume that formula's antecedents:  
(4)   
m is square
 
    and also:  
(5)   
n is square
 
      and we'll work on proving the consequent:
(6)     
mn is square
          ≡      by (1) with "a := mn"
(7)     
∃b((mn) = b²)
      Removing that variable quantification
clarifies that to prove that formula,
it will suffice to find a value for the variable.
(8)     
(mn) = b²
    Applying the equivalence in formula (1)
to formula (4)
with "a := m"
yields: 		 
(9)   
∃b(m = b²)
 
    That formula indicates there exists
some value say "s" satisfying that
quantified formula, i.e.: 		 
(10)   
m = s²
 
    Applying the equivalence in formula (1)
to formula (5)
with "a := n"
yields: 		 
(11)   
∃b(n = b²)
 
    That formula indicates there exists
some value say "t" satisfying that
quantified formula, i.e.: 		 
(12)   
n = t²
 
       
      We'll work on transforming the left-hand side of formula (8)
to the right-hand side as follows:
(13)     
mn
          =      by (10)
(14)     
s²n
          =      by (12)
(15)     
s²t²
          =      by (2) with "x := s"
(16)     
[s·s]t²
          =      by (2) with "x := t"
(17)     
[s·s][t·t]
          =      by basic algebra/arithmetic
(18)     
[s·t][s·t]
          =      by (2) with "x := s·t"
(19)     
(s·t)²
      And that satisfies our earlier goal (8)
with "b := s·t"; i.e. we have:
(20)     
true
      That concludes this part of the proof.
       
      Thus, the theorem that was given is true.

identification:
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1233599154843
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