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 → (φ → ∀x(x = y → φ))))

Assertion

Ref	Expression
ax11indn	⊢ (¬ ∀x x = y → (x = y → (¬ φ → ∀x(x = y → ¬ φ))))

Proof of Theorem ax11indn

Step	Hyp	Ref	Expression
1		19.8a 1756	. . 3 ⊢ ((x = y ∧ ¬ φ) → ∃x(x = y ∧ ¬ φ))
2		exanali 1585	. . . 4 ⊢ (∃x(x = y ∧ ¬ φ) ↔ ¬ ∀x(x = y → φ))
3		hbn1 1730	. . . . 5 ⊢ (¬ ∀x x = y → ∀x ¬ ∀x x = y)
4		hbn1 1730	. . . . 5 ⊢ (¬ ∀x(x = y → φ) → ∀x ¬ ∀x(x = y → φ))
5		ax11indn.1	. . . . . . 7 ⊢ (¬ ∀x x = y → (x = y → (φ → ∀x(x = y → φ))))
6		con3 126	. . . . . . 7 ⊢ ((φ → ∀x(x = y → φ)) → (¬ ∀x(x = y → φ) → ¬ φ))
7	5, 6	syl6 29	. . . . . 6 ⊢ (¬ ∀x x = y → (x = y → (¬ ∀x(x = y → φ) → ¬ φ)))
8	7	com23 72	. . . . 5 ⊢ (¬ ∀x x = y → (¬ ∀x(x = y → φ) → (x = y → ¬ φ)))
9	3, 4, 8	alrimdh 1587	. . . 4 ⊢ (¬ ∀x x = y → (¬ ∀x(x = y → φ) → ∀x(x = y → ¬ φ)))
10	2, 9	syl5bi 208	. . 3 ⊢ (¬ ∀x x = y → (∃x(x = y ∧ ¬ φ) → ∀x(x = y → ¬ φ)))
11	1, 10	syl5 28	. 2 ⊢ (¬ ∀x x = y → ((x = y ∧ ¬ φ) → ∀x(x = y → ¬ φ)))
12	11	exp3a 425	1 ⊢ (¬ ∀x x = y → (x = y → (¬ φ → ∀x(x = y → ¬ φ))))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ∧ wa 358  ∀wal 1540  ∃wex 1541

This proof depends on 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 proof depends on definitions:  df-bi 177  df-an 360  df-ex 1542

This theorem is used by:  ax11indi  2196