Theorem sbfal 35387

Description: Substitution does not change falsity. (Contributed by Giovanni Mascellani, 24-May-2019.)

Hypothesis

Ref	Expression
sbfal.1	⊢ 𝐴 ∈ V

Assertion

Ref	Expression
sbfal	⊢ ([𝐴 / 𝑥]⊥ ↔ ⊥)

Proof of Theorem sbfal

Step	Hyp	Ref	Expression
1		sbfal.1	. 2 ⊢ 𝐴 ∈ V
2		sbcg 3849	. 2 ⊢ (𝐴 ∈ V → ([𝐴 / 𝑥]⊥ ↔ ⊥))
3	1, 2	ax-mp 5	1 ⊢ ([𝐴 / 𝑥]⊥ ↔ ⊥)

Syntax hints:   ↔ wb 208  ⊥wfal 1549   ∈ wcel 2114  Vcvv 3496  [wsbc 3774

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1970  ax-7 2015  ax-8 2116  ax-9 2124  ax-12 2177  ax-ext 2795

This theorem depends on definitions:  df-bi 209  df-an 399  df-ex 1781  df-nf 1785  df-sb 2070  df-clab 2802  df-cleq 2816  df-clel 2895  df-sbc 3775

This theorem is referenced by: (None)