Theorem spesbcd 3834

Description: form of spsbc 3755. (Contributed by Mario Carneiro, 9-Feb-2017.)

Hypothesis

Ref	Expression
spesbcd.1	⊢ (𝜑 → [𝐴 / 𝑥]𝜓)

Assertion

Ref	Expression
spesbcd	⊢ (𝜑 → ∃𝑥𝜓)

Proof of Theorem spesbcd

Step	Hyp	Ref	Expression
1		spesbcd.1	. 2 ⊢ (𝜑 → [𝐴 / 𝑥]𝜓)
2		spesbc 3833	. 2 ⊢ ([𝐴 / 𝑥]𝜓 → ∃𝑥𝜓)
3	1, 2	syl 18	1 ⊢ (𝜑 → ∃𝑥𝜓)

This proof depends on syntax axioms:   → wi 4  ∃wex 1812  [wsbc 3742

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 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-10 2180  ax-11 2197  ax-12 2216  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-ex 1813  df-nf 1817  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-nfc 2913  df-ral 3081  df-rex 3091  df-v 3456  df-sbc 3743

This theorem is used by:  euotd  5493  ex-natded9.26  31077  bnj1465  35586  spesbcdi  39318  iscard4  44795  minregex  44796  brtrclfv2  44988  cotrclrcl  45003  rspesbcd  46180