Metamath Proof Explorer


Theorem csbcnv

Description: Move class substitution in and out of the converse of a relation. Version of csbcnvgALT without a sethood antecedent but depending on more axioms. (Contributed by Thierry Arnoux, 8-Feb-2017) (Revised by NM, 23-Aug-2018)

Ref Expression
Assertion csbcnv 𝐴 / 𝑥 𝐹 = 𝐴 / 𝑥 𝐹

Proof

Step Hyp Ref Expression
1 sbcbr ( [ 𝐴 / 𝑥 ] 𝑧 𝐹 𝑦𝑧 𝐴 / 𝑥 𝐹 𝑦 )
2 1 opabbii { ⟨ 𝑦 , 𝑧 ⟩ ∣ [ 𝐴 / 𝑥 ] 𝑧 𝐹 𝑦 } = { ⟨ 𝑦 , 𝑧 ⟩ ∣ 𝑧 𝐴 / 𝑥 𝐹 𝑦 }
3 csbopab 𝐴 / 𝑥 { ⟨ 𝑦 , 𝑧 ⟩ ∣ 𝑧 𝐹 𝑦 } = { ⟨ 𝑦 , 𝑧 ⟩ ∣ [ 𝐴 / 𝑥 ] 𝑧 𝐹 𝑦 }
4 df-cnv 𝐴 / 𝑥 𝐹 = { ⟨ 𝑦 , 𝑧 ⟩ ∣ 𝑧 𝐴 / 𝑥 𝐹 𝑦 }
5 2 3 4 3eqtr4ri 𝐴 / 𝑥 𝐹 = 𝐴 / 𝑥 { ⟨ 𝑦 , 𝑧 ⟩ ∣ 𝑧 𝐹 𝑦 }
6 df-cnv 𝐹 = { ⟨ 𝑦 , 𝑧 ⟩ ∣ 𝑧 𝐹 𝑦 }
7 6 csbeq2i 𝐴 / 𝑥 𝐹 = 𝐴 / 𝑥 { ⟨ 𝑦 , 𝑧 ⟩ ∣ 𝑧 𝐹 𝑦 }
8 5 7 eqtr4i 𝐴 / 𝑥 𝐹 = 𝐴 / 𝑥 𝐹