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Theorem i5con 272
 Description: Converse of →5 .
Assertion
Ref Expression
i5con (a5 b) = (b5 a )

Proof of Theorem i5con
StepHypRef Expression
1 ancom 74 . . . 4 (ab ) = (ba )
2 ax-a2 31 . . . . 5 ((ab) ∪ (ab)) = ((ab) ∪ (ab))
3 ancom 74 . . . . . . 7 (ab) = (ba )
4 ax-a1 30 . . . . . . . 8 b = b
54ran 78 . . . . . . 7 (ba ) = (b a )
63, 5ax-r2 36 . . . . . 6 (ab) = (b a )
7 ancom 74 . . . . . . 7 (ab) = (ba)
8 ax-a1 30 . . . . . . . 8 a = a
94, 82an 79 . . . . . . 7 (ba) = (b a )
107, 9ax-r2 36 . . . . . 6 (ab) = (b a )
116, 102or 72 . . . . 5 ((ab) ∪ (ab)) = ((b a ) ∪ (b a ))
122, 11ax-r2 36 . . . 4 ((ab) ∪ (ab)) = ((b a ) ∪ (b a ))
131, 122or 72 . . 3 ((ab ) ∪ ((ab) ∪ (ab))) = ((ba ) ∪ ((b a ) ∪ (b a )))
14 ax-a2 31 . . 3 (((ab) ∪ (ab)) ∪ (ab )) = ((ab ) ∪ ((ab) ∪ (ab)))
15 ax-a3 32 . . 3 (((ba ) ∪ (b a )) ∪ (b a )) = ((ba ) ∪ ((b a ) ∪ (b a )))
1613, 14, 153tr1 63 . 2 (((ab) ∪ (ab)) ∪ (ab )) = (((ba ) ∪ (b a )) ∪ (b a ))
17 df-i5 48 . 2 (a5 b) = (((ab) ∪ (ab)) ∪ (ab ))
18 df-i5 48 . 2 (b5 a ) = (((ba ) ∪ (b a )) ∪ (b a ))
1916, 17, 183tr1 63 1 (a5 b) = (b5 a )
 Colors of variables: term Syntax hints:   = wb 1  ⊥ wn 4   ∪ wo 6   ∩ wa 7   →5 wi5 16 This theorem was proved from axioms:  ax-a1 30  ax-a2 31  ax-a3 32  ax-r1 35  ax-r2 36  ax-r4 37  ax-r5 38 This theorem depends on definitions:  df-a 40  df-i5 48 This theorem is referenced by:  nom45  330
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