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Theorem ax12indn 34751
 Description: Induction step for constructing a substitution instance of ax-c15 34697 without using ax-c15 34697. Negation case. (Contributed by NM, 21-Jan-2007.) (Proof modification is discouraged.) (New usage is discouraged.)
Hypothesis
Ref Expression
ax12indn.1 (¬ ∀𝑥 𝑥 = 𝑦 → (𝑥 = 𝑦 → (𝜑 → ∀𝑥(𝑥 = 𝑦𝜑))))
Assertion
Ref Expression
ax12indn (¬ ∀𝑥 𝑥 = 𝑦 → (𝑥 = 𝑦 → (¬ 𝜑 → ∀𝑥(𝑥 = 𝑦 → ¬ 𝜑))))

Proof of Theorem ax12indn
StepHypRef Expression
1 19.8a 2206 . . 3 ((𝑥 = 𝑦 ∧ ¬ 𝜑) → ∃𝑥(𝑥 = 𝑦 ∧ ¬ 𝜑))
2 exanali 1937 . . . 4 (∃𝑥(𝑥 = 𝑦 ∧ ¬ 𝜑) ↔ ¬ ∀𝑥(𝑥 = 𝑦𝜑))
3 hbn1 2175 . . . . 5 (¬ ∀𝑥 𝑥 = 𝑦 → ∀𝑥 ¬ ∀𝑥 𝑥 = 𝑦)
4 hbn1 2175 . . . . 5 (¬ ∀𝑥(𝑥 = 𝑦𝜑) → ∀𝑥 ¬ ∀𝑥(𝑥 = 𝑦𝜑))
5 ax12indn.1 . . . . . . 7 (¬ ∀𝑥 𝑥 = 𝑦 → (𝑥 = 𝑦 → (𝜑 → ∀𝑥(𝑥 = 𝑦𝜑))))
6 con3 150 . . . . . . 7 ((𝜑 → ∀𝑥(𝑥 = 𝑦𝜑)) → (¬ ∀𝑥(𝑥 = 𝑦𝜑) → ¬ 𝜑))
75, 6syl6 35 . . . . . 6 (¬ ∀𝑥 𝑥 = 𝑦 → (𝑥 = 𝑦 → (¬ ∀𝑥(𝑥 = 𝑦𝜑) → ¬ 𝜑)))
87com23 86 . . . . 5 (¬ ∀𝑥 𝑥 = 𝑦 → (¬ ∀𝑥(𝑥 = 𝑦𝜑) → (𝑥 = 𝑦 → ¬ 𝜑)))
93, 4, 8alrimdh 1941 . . . 4 (¬ ∀𝑥 𝑥 = 𝑦 → (¬ ∀𝑥(𝑥 = 𝑦𝜑) → ∀𝑥(𝑥 = 𝑦 → ¬ 𝜑)))
102, 9syl5bi 232 . . 3 (¬ ∀𝑥 𝑥 = 𝑦 → (∃𝑥(𝑥 = 𝑦 ∧ ¬ 𝜑) → ∀𝑥(𝑥 = 𝑦 → ¬ 𝜑)))
111, 10syl5 34 . 2 (¬ ∀𝑥 𝑥 = 𝑦 → ((𝑥 = 𝑦 ∧ ¬ 𝜑) → ∀𝑥(𝑥 = 𝑦 → ¬ 𝜑)))
1211expd 400 1 (¬ ∀𝑥 𝑥 = 𝑦 → (𝑥 = 𝑦 → (¬ 𝜑 → ∀𝑥(𝑥 = 𝑦 → ¬ 𝜑))))
 Colors of variables: wff setvar class Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 382  ∀wal 1629  ∃wex 1852 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1870  ax-4 1885  ax-5 1991  ax-6 2057  ax-7 2093  ax-10 2174  ax-12 2203 This theorem depends on definitions:  df-bi 197  df-an 383  df-ex 1853 This theorem is referenced by:  ax12indi  34752
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