Metamath Proof Explorer


Theorem sbceqbii

Description: Formula-building inference for class substitution. General version of sbcbii . (Contributed by GG, 1-Sep-2025)

Ref Expression
Hypotheses sbceqbii.1 ⊢ 𝐴 = 𝐵
sbceqbii.2 ⊢ ( 𝜑 ↔ 𝜓 )
Assertion sbceqbii ( [ 𝐴 / 𝑥 ] 𝜑 ↔ [ 𝐵 / 𝑥 ] 𝜓 )

Proof

Step Hyp Ref Expression
1 sbceqbii.1 ⊢ 𝐴 = 𝐵
2 sbceqbii.2 ⊢ ( 𝜑 ↔ 𝜓 )
3 2 abbii ⊢ { 𝑥 ∣ 𝜑 } = { 𝑥 ∣ 𝜓 }
4 1 3 eleq12i ⊢ ( 𝐴 ∈ { 𝑥 ∣ 𝜑 } ↔ 𝐵 ∈ { 𝑥 ∣ 𝜓 } )
5 df-sbc ⊢ ( [ 𝐴 / 𝑥 ] 𝜑 ↔ 𝐴 ∈ { 𝑥 ∣ 𝜑 } )
6 df-sbc ⊢ ( [ 𝐵 / 𝑥 ] 𝜓 ↔ 𝐵 ∈ { 𝑥 ∣ 𝜓 } )
7 4 5 6 3bitr4i ⊢ ( [ 𝐴 / 𝑥 ] 𝜑 ↔ [ 𝐵 / 𝑥 ] 𝜓 )