Metamath Proof Explorer


Theorem sbciegOLD

Description: Obsolete version of sbcieg as of 12-Oct-2024. (Contributed by NM, 10-Nov-2005) (Proof modification is discouraged.) (New usage is discouraged.)

Ref Expression
Hypothesis sbciegOLD.1 ( 𝑥 = 𝐴 → ( 𝜑𝜓 ) )
Assertion sbciegOLD ( 𝐴𝑉 → ( [ 𝐴 / 𝑥 ] 𝜑𝜓 ) )

Proof

Step Hyp Ref Expression
1 sbciegOLD.1 ( 𝑥 = 𝐴 → ( 𝜑𝜓 ) )
2 nfv 𝑥 𝜓
3 2 1 sbciegf ( 𝐴𝑉 → ( [ 𝐴 / 𝑥 ] 𝜑𝜓 ) )