Metamath Proof Explorer


Theorem sb6x

Description: Equivalence involving substitution for a variable not free. Usage of this theorem is discouraged because it depends on ax-13 . Usage of sb6 is preferred, which requires fewer axioms. (Contributed by NM, 2-Jun-1993) (Revised by Mario Carneiro, 4-Oct-2016) (New usage is discouraged.)

Ref Expression
Hypothesis sb6x.1 𝑥 𝜑
Assertion sb6x ( [ 𝑦 / 𝑥 ] 𝜑 ↔ ∀ 𝑥 ( 𝑥 = 𝑦𝜑 ) )

Proof

Step Hyp Ref Expression
1 sb6x.1 𝑥 𝜑
2 1 sbf ( [ 𝑦 / 𝑥 ] 𝜑𝜑 )
3 biidd ( 𝑥 = 𝑦 → ( 𝜑𝜑 ) )
4 1 3 equsal ( ∀ 𝑥 ( 𝑥 = 𝑦𝜑 ) ↔ 𝜑 )
5 2 4 bitr4i ( [ 𝑦 / 𝑥 ] 𝜑 ↔ ∀ 𝑥 ( 𝑥 = 𝑦𝜑 ) )