Theorem spesbcd

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

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

Proof

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