Metamath Proof Explorer


Theorem sbequ8

Description: Elimination of equality from antecedent after substitution. Usage of this theorem is discouraged because it depends on ax-13 . (Contributed by NM, 5-Aug-1993) Reduce dependencies on axioms. (Revised by Wolf Lammen, 28-Jul-2018) Revise df-sb . (Revised by Wolf Lammen, 28-Jul-2023) (New usage is discouraged.)

Ref Expression
Assertion sbequ8 ( [ 𝑦 / 𝑥 ] 𝜑 ↔ [ 𝑦 / 𝑥 ] ( 𝑥 = 𝑦𝜑 ) )

Proof

Step Hyp Ref Expression
1 equsb1 [ 𝑦 / 𝑥 ] 𝑥 = 𝑦
2 1 a1bi ( [ 𝑦 / 𝑥 ] 𝜑 ↔ ( [ 𝑦 / 𝑥 ] 𝑥 = 𝑦 → [ 𝑦 / 𝑥 ] 𝜑 ) )
3 sbim ( [ 𝑦 / 𝑥 ] ( 𝑥 = 𝑦𝜑 ) ↔ ( [ 𝑦 / 𝑥 ] 𝑥 = 𝑦 → [ 𝑦 / 𝑥 ] 𝜑 ) )
4 2 3 bitr4i ( [ 𝑦 / 𝑥 ] 𝜑 ↔ [ 𝑦 / 𝑥 ] ( 𝑥 = 𝑦𝜑 ) )