Metamath Proof Explorer


Theorem sbhb

Description: Two ways of expressing " x is (effectively) not free in ph ". Usage of this theorem is discouraged because it depends on ax-13 . (Contributed by NM, 29-May-2009) (New usage is discouraged.)

Ref Expression
Assertion sbhb ( ( 𝜑 → ∀ 𝑥 𝜑 ) ↔ ∀ 𝑦 ( 𝜑 → [ 𝑦 / 𝑥 ] 𝜑 ) )

Proof

Step Hyp Ref Expression
1 nfv 𝑦 𝜑
2 1 sb8 ( ∀ 𝑥 𝜑 ↔ ∀ 𝑦 [ 𝑦 / 𝑥 ] 𝜑 )
3 2 imbi2i ( ( 𝜑 → ∀ 𝑥 𝜑 ) ↔ ( 𝜑 → ∀ 𝑦 [ 𝑦 / 𝑥 ] 𝜑 ) )
4 19.21v ( ∀ 𝑦 ( 𝜑 → [ 𝑦 / 𝑥 ] 𝜑 ) ↔ ( 𝜑 → ∀ 𝑦 [ 𝑦 / 𝑥 ] 𝜑 ) )
5 3 4 bitr4i ( ( 𝜑 → ∀ 𝑥 𝜑 ) ↔ ∀ 𝑦 ( 𝜑 → [ 𝑦 / 𝑥 ] 𝜑 ) )