Theorem ax11indn 1366

Description: Induction step for constructing a substitution instance of ax-11o 1218 without using ax-11o 1218. Negation case.

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		hbn1 1015	. . . . 5 ⊢ (¬ ∀x x = y → ∀x ¬ ∀x x = y)
2		hbn1 1015	. . . . 5 ⊢ (¬ ∀x(x = y → φ) → ∀x ¬ ∀x(x = y → φ))
3		ax11indn.1	. . . . . . 7 ⊢ (¬ ∀x x = y → (x = y → (φ → ∀x(x = y → φ))))
4		con3 94	. . . . . . 7 ⊢ ((φ → ∀x(x = y → φ)) → (¬ ∀x(x = y → φ) → ¬ φ))
5	3, 4	syl6 22	. . . . . 6 ⊢ (¬ ∀x x = y → (x = y → (¬ ∀x(x = y → φ) → ¬ φ)))
6	5	com23 32	. . . . 5 ⊢ (¬ ∀x x = y → (¬ ∀x(x = y → φ) → (x = y → ¬ φ)))
7	1, 2, 6	19.21ad 1059	. . . 4 ⊢ (¬ ∀x x = y → (¬ ∀x(x = y → φ) → ∀x(x = y → ¬ φ)))
8		exanali 1043	. . . 4 ⊢ (∃x(x = y ⋀ ¬ φ) ↔ ¬ ∀x(x = y → φ))
9	7, 8	syl5ib 206	. . 3 ⊢ (¬ ∀x x = y → (∃x(x = y ⋀ ¬ φ) → ∀x(x = y → ¬ φ)))
10		19.8a 1029	. . 3 ⊢ ((x = y ⋀ ¬ φ) → ∃x(x = y ⋀ ¬ φ))
11	9, 10	syl5 21	. 2 ⊢ (¬ ∀x x = y → ((x = y ⋀ ¬ φ) → ∀x(x = y → ¬ φ)))
12	11	exp3a 375	1 ⊢ (¬ ∀x x = y → (x = y → (¬ φ → ∀x(x = y → ¬ φ))))

Syntax hints:  ¬ wn 2   → wi 3   ⋀ wa 223  ∀wal 954   = wceq 956  ∃wex 980

This theorem is referenced by:  ax11indi 1367  a12studyALT 1379

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-gen 963  ax-4 973  ax-5o 975  ax-6o 978

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 981

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