S_union_empty_equals_S

mcguireh

Proof of the following formula: 

(∀S)[S ∪ ∅ = S]

(#)   Suppositions and derivations Theorem and subgoals
    We presuppose the following:  
(1)   
∀S1∀S2
  [S1 = S2
     ↔
   (∀x)
     (x ∈ S1
        ↔
      x ∈ S2)]
 
(2)   
∀S1∀S2∀x
   [x ∈ S1 ∪ S2
     ↔
   (x ∈ S1  ∨  x ∈ S2)]
 
       
      We start working with the formula being proved
as follows:
(3)     
(∀S)[S ∪ ∅ = S]
      Well, let an arbitrary value "A" be given;
then we should prove the following:
(4)     
A ∪ ∅ = A
          ≡      by (1) with [S1 := A∪∅, S2 := A]
(5)     
(∀x)(x ∈ (A ∪ ∅)  ↔  x ∈ A)
      Well, let an arbitrary value "r" be given;
then we should prove the following:
(6)     
r ∈ (A ∪ ∅)  ↔  r ∈ A
       
      We'll work on transforming the left-hand side of formula (6)
to the right-hand side as follows:
(7)     
r ∈ (A ∪ ∅)
          ≡      by (2) with [x := r, S1 := A, S2 := ∅]
(8)     
r ∈ A  ∨  r ∈ ∅
          ≡      by simplifying
(9)     
r ∈ A
      And that satisfies our earlier goal (6).
       
      That sequence of transformations shows that (6) is:
(10)     
true
      That concludes the proof.

identification:
1256648743414
1256648544257
199157