Theorem als1d 17300

Description: Deduction rule: Given "all some" applied to a top-level inference, you can extract the "for all" part. (Contributed by David A. Wheeler, 20-Oct-2018.)

Hypothesis

Ref	Expression
als1d.1	⊢ (𝜑 → ∀∃𝑥(𝜓 → 𝜒))

Assertion

Ref	Expression
als1d	⊢ (𝜑 → ∀𝑥(𝜓 → 𝜒))

Proof of Theorem als1d

Step	Hyp	Ref	Expression
1		als1d.1	. . 3 ⊢ (𝜑 → ∀∃𝑥(𝜓 → 𝜒))
2		df-als 17295	. . 3 ⊢ (∀∃𝑥(𝜓 → 𝜒) ↔ (∀𝑥(𝜓 → 𝜒) ∧ ∃𝑥𝜓))
3	1, 2	sylib 122	. 2 ⊢ (𝜑 → (∀𝑥(𝜓 → 𝜒) ∧ ∃𝑥𝜓))
4	3	simpld 112	1 ⊢ (𝜑 → ∀𝑥(𝜓 → 𝜒))

This proof depends on syntax axioms:   → wi 4   ∧ wa 104  ∀wal 1400  ∃wex 1545  ∀∃wals 17293

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106

This proof depends on definitions:  df-bi 117  df-als 17295

This theorem is used by: (None)