Theorem ax12inda2 40098

Description: Induction step for constructing a substitution instance of ax-c15 40040 without using ax-c15 40040. Quantification case. When 𝑧 and 𝑦 are distinct, this theorem avoids the dummy variables needed by the more general ax12inda 40099. (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 40085	. . . . 5 ⊢ (∀𝑦 𝑦 = 𝑧 → ((𝑥 = 𝑦 → ∀𝑧𝜑) → ∀𝑥(𝑥 = 𝑦 → ∀𝑧𝜑)))
3	1, 2	syl5 35	. . . 4 ⊢ (∀𝑦 𝑦 = 𝑧 → (∀𝑧𝜑 → ∀𝑥(𝑥 = 𝑦 → ∀𝑧𝜑)))
4	3	a1d 26	. . 3 ⊢ (∀𝑦 𝑦 = 𝑧 → (𝑥 = 𝑦 → (∀𝑧𝜑 → ∀𝑥(𝑥 = 𝑦 → ∀𝑧𝜑))))
5	4	a1d 26	. 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 40096	. 2 ⊢ (¬ ∀𝑦 𝑦 = 𝑧 → (¬ ∀𝑥 𝑥 = 𝑦 → (𝑥 = 𝑦 → (∀𝑧𝜑 → ∀𝑥(𝑥 = 𝑦 → ∀𝑧𝜑)))))
8	5, 7	pm2.61i 184	1 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → (𝑥 = 𝑦 → (∀𝑧𝜑 → ∀𝑥(𝑥 = 𝑦 → ∀𝑧𝜑))))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4  ∀wal 1567

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-10 2177  ax-11 2193  ax-12 2212  ax-13 2401  ax-c5 40034  ax-c4 40035  ax-c7 40036  ax-c10 40037  ax-c11 40038  ax-c9 40041  ax-c16 40043

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1572  df-ex 1812  df-nf 1816

This theorem is used by:  ax12inda  40099