Theorem spd 50497

Description: Specialization deduction, using implicit substitution. Based on the proof of spimed 2423. (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 2418	. . . 4 ⊢ ∃𝑥 𝑥 = 𝑦
2		spd.2	. . . . 5 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
3	2	biimpd 232	. . . 4 ⊢ (𝑥 = 𝑦 → (𝜑 → 𝜓))
4	1, 3	eximii 1870	. . 3 ⊢ ∃𝑥(𝜑 → 𝜓)
5	4	19.35i 1911	. 2 ⊢ (∀𝑥𝜑 → ∃𝑥𝜓)
6		spd.1	. . 3 ⊢ (𝜒 → Ⅎ𝑥𝜓)
7	6	19.9d 2242	. 2 ⊢ (𝜒 → (∃𝑥𝜓 → 𝜓))
8	5, 7	syl5 35	1 ⊢ (𝜒 → (∀𝑥𝜑 → 𝜓))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209  ∀wal 1568  ∃wex 1812  Ⅎwnf 1816

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-12 2216  ax-13 2407

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-nf 1817

This theorem is used by: (None)