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Theorem ax11indn 2195
 Description: Induction step for constructing a substitution instance of ax-11o 2141 without using ax-11o 2141. Negation case. (Contributed by NM, 21-Jan-2007.) (Proof modification is discouraged.) (New usage is discouraged.)
Hypothesis
Ref Expression
ax11indn.1 x x = y → (x = y → (φx(x = yφ))))
Assertion
Ref Expression
ax11indn x x = y → (x = y → (¬ φx(x = y → ¬ φ))))

Proof of Theorem ax11indn
StepHypRef Expression
1 19.8a 1756 . . 3 ((x = y ¬ φ) → x(x = y ¬ φ))
2 exanali 1585 . . . 4 (x(x = y ¬ φ) ↔ ¬ x(x = yφ))
3 hbn1 1730 . . . . 5 x x = yx ¬ x x = y)
4 hbn1 1730 . . . . 5 x(x = yφ) → x ¬ x(x = yφ))
5 ax11indn.1 . . . . . . 7 x x = y → (x = y → (φx(x = yφ))))
6 con3 126 . . . . . . 7 ((φx(x = yφ)) → (¬ x(x = yφ) → ¬ φ))
75, 6syl6 29 . . . . . 6 x x = y → (x = y → (¬ x(x = yφ) → ¬ φ)))
87com23 72 . . . . 5 x x = y → (¬ x(x = yφ) → (x = y → ¬ φ)))
93, 4, 8alrimdh 1587 . . . 4 x x = y → (¬ x(x = yφ) → x(x = y → ¬ φ)))
102, 9syl5bi 208 . . 3 x x = y → (x(x = y ¬ φ) → x(x = y → ¬ φ)))
111, 10syl5 28 . 2 x x = y → ((x = y ¬ φ) → x(x = y → ¬ φ)))
1211exp3a 425 1 x x = y → (x = y → (¬ φx(x = y → ¬ φ))))
 Colors of variables: wff setvar class Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 358  ∀wal 1540  ∃wex 1541 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-11 1746 This theorem depends on definitions:  df-bi 177  df-an 360  df-ex 1542 This theorem is referenced by:  ax11indi  2196
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