Theorem spesbcd 3837

Description: form of spsbc 3759. (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 3836	. 2 ⊢ ([𝐴 / 𝑥]𝜓 → ∃𝑥𝜓)
3	1, 2	syl 18	1 ⊢ (𝜑 → ∃𝑥𝜓)

This proof depends on syntax axioms:   → wi 4  ∃wex 1812  [wsbc 3746

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-8 2148  ax-9 2156  ax-10 2179  ax-11 2195  ax-12 2216  ax-ext 2737

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 2100  df-clab 2744  df-cleq 2757  df-clel 2840  df-nfc 2914  df-ral 3082  df-rex 3092  df-v 3459  df-sbc 3747

This theorem is used by:  euotd  5498  ex-natded9.26  30841  bnj1465  35298  spesbcdi  38827  iscard4  44317  minregex  44318  brtrclfv2  44511  cotrclrcl  44526  rspesbcd  45704