Theorem spd 46270

Description: Specialization deduction, using implicit substitution. Based on the proof of spimed 2388. (Contributed by Emmett Weisz, 17-Jan-2020.)

Hypotheses

Ref	Expression
spd.1	⊢ (𝜒 → Ⅎ𝑥𝜓)
spd.2	⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))

Assertion

Ref	Expression
spd	⊢ (𝜒 → (∀𝑥𝜑 → 𝜓))

Proof of Theorem spd

Step	Hyp	Ref	Expression
1		ax6e 2383	. . . 4 ⊢ ∃𝑥 𝑥 = 𝑦
2		spd.2	. . . . 5 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
3	2	biimpd 228	. . . 4 ⊢ (𝑥 = 𝑦 → (𝜑 → 𝜓))
4	1, 3	eximii 1840	. . 3 ⊢ ∃𝑥(𝜑 → 𝜓)
5	4	19.35i 1882	. 2 ⊢ (∀𝑥𝜑 → ∃𝑥𝜓)
6		spd.1	. . 3 ⊢ (𝜒 → Ⅎ𝑥𝜓)
7	6	19.9d 2199	. 2 ⊢ (𝜒 → (∃𝑥𝜓 → 𝜓))
8	5, 7	syl5 34	1 ⊢ (𝜒 → (∀𝑥𝜑 → 𝜓))

Syntax hints:   → wi 4   ↔ wb 205  ∀wal 1537  ∃wex 1783  Ⅎwnf 1787

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-12 2173  ax-13 2372

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

This theorem is referenced by: (None)