Theorem bj-sbcex

Description: Proof of sbcex when taking bj-df-sb as definition. (Contributed by BJ, 19-Feb-2026) (Proof modification is discouraged.) (New usage is discouraged.)

	Ref	Expression
Assertion	bj-sbcex	⊢ ( [ 𝐴 / 𝑥 ] 𝜑 → 𝐴 ∈ V )

Proof

Step	Hyp	Ref	Expression
1		exsimpl	⊢ ( ∃ 𝑦 ( 𝑦 = 𝐴 ∧ ∀ 𝑥 ( 𝑥 = 𝑦 → 𝜑 ) ) → ∃ 𝑦 𝑦 = 𝐴 )
2		bj-df-sb	⊢ ( [ 𝐴 / 𝑥 ] 𝜑 ↔ ∃ 𝑦 ( 𝑦 = 𝐴 ∧ ∀ 𝑥 ( 𝑥 = 𝑦 → 𝜑 ) ) )
3		isset	⊢ ( 𝐴 ∈ V ↔ ∃ 𝑦 𝑦 = 𝐴 )
4	1 2 3	3imtr4i	⊢ ( [ 𝐴 / 𝑥 ] 𝜑 → 𝐴 ∈ V )