Theorem ax12inda2 36888

Description: Induction step for constructing a substitution instance of ax-c15 36830 without using ax-c15 36830. Quantification case. When 𝑧 and 𝑦 are distinct, this theorem avoids the dummy variables needed by the more general ax12inda 36889. (Contributed by NM, 24-Jan-2007.) (Proof modification is discouraged.) (New usage is discouraged.)

Hypothesis

Ref	Expression
ax12inda2.1	⊢ (¬ ∀𝑥 𝑥 = 𝑦 → (𝑥 = 𝑦 → (𝜑 → ∀𝑥(𝑥 = 𝑦 → 𝜑))))

Assertion

Ref	Expression
ax12inda2	⊢ (¬ ∀𝑥 𝑥 = 𝑦 → (𝑥 = 𝑦 → (∀𝑧𝜑 → ∀𝑥(𝑥 = 𝑦 → ∀𝑧𝜑))))

Distinct variable group:   𝑦,𝑧

Allowed substitution hints:   𝜑(𝑥,𝑦,𝑧)

Proof of Theorem ax12inda2

Step	Hyp	Ref	Expression
1		ax-1 6	. . . . 5 ⊢ (∀𝑧𝜑 → (𝑥 = 𝑦 → ∀𝑧𝜑))
2		axc16g-o 36875	. . . . 5 ⊢ (∀𝑦 𝑦 = 𝑧 → ((𝑥 = 𝑦 → ∀𝑧𝜑) → ∀𝑥(𝑥 = 𝑦 → ∀𝑧𝜑)))
3	1, 2	syl5 34	. . . 4 ⊢ (∀𝑦 𝑦 = 𝑧 → (∀𝑧𝜑 → ∀𝑥(𝑥 = 𝑦 → ∀𝑧𝜑)))
4	3	a1d 25	. . 3 ⊢ (∀𝑦 𝑦 = 𝑧 → (𝑥 = 𝑦 → (∀𝑧𝜑 → ∀𝑥(𝑥 = 𝑦 → ∀𝑧𝜑))))
5	4	a1d 25	. 2 ⊢ (∀𝑦 𝑦 = 𝑧 → (¬ ∀𝑥 𝑥 = 𝑦 → (𝑥 = 𝑦 → (∀𝑧𝜑 → ∀𝑥(𝑥 = 𝑦 → ∀𝑧𝜑)))))
6		ax12inda2.1	. . 3 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → (𝑥 = 𝑦 → (𝜑 → ∀𝑥(𝑥 = 𝑦 → 𝜑))))
7	6	ax12indalem 36886	. 2 ⊢ (¬ ∀𝑦 𝑦 = 𝑧 → (¬ ∀𝑥 𝑥 = 𝑦 → (𝑥 = 𝑦 → (∀𝑧𝜑 → ∀𝑥(𝑥 = 𝑦 → ∀𝑧𝜑)))))
8	5, 7	pm2.61i 182	1 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → (𝑥 = 𝑦 → (∀𝑧𝜑 → ∀𝑥(𝑥 = 𝑦 → ∀𝑧𝜑))))

Syntax hints:  ¬ wn 3   → wi 4  ∀wal 1537

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1799  ax-4 1813  ax-5 1914  ax-6 1972  ax-7 2012  ax-10 2139  ax-11 2156  ax-12 2173  ax-13 2372  ax-c5 36824  ax-c4 36825  ax-c7 36826  ax-c10 36827  ax-c11 36828  ax-c9 36831  ax-c16 36833

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 844  df-tru 1542  df-ex 1784  df-nf 1788

This theorem is referenced by:  ax12inda  36889