Theorem elnelall 2415

Description: A contradiction concerning membership implies anything. (Contributed by Alexander van der Vekens, 25-Jan-2018.)

Assertion

Ref	Expression
elnelall	⊢ (𝐴 ∈ 𝐵 → (𝐴 ∉ 𝐵 → 𝜑))

Proof of Theorem elnelall

Step	Hyp	Ref	Expression
1		df-nel 2404	. 2 ⊢ (𝐴 ∉ 𝐵 ↔ ¬ 𝐴 ∈ 𝐵)
2		pm2.24 610	. 2 ⊢ (𝐴 ∈ 𝐵 → (¬ 𝐴 ∈ 𝐵 → 𝜑))
3	1, 2	syl5bi 151	1 ⊢ (𝐴 ∈ 𝐵 → (𝐴 ∉ 𝐵 → 𝜑))

Syntax hints:  ¬ wn 3   → wi 4   ∈ wcel 1480   ∉ wnel 2403

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-in2 604

This theorem depends on definitions:  df-bi 116  df-nel 2404

This theorem is referenced by:  xnn0lenn0nn0  9648