Metamath Proof Explorer


Theorem csbcnvgALT

Description: Move class substitution in and out of the converse of a relation. Version of csbcnv with a sethood antecedent but depending on fewer axioms. (Contributed by Thierry Arnoux, 8-Feb-2017) (New usage is discouraged.) (Proof modification is discouraged.)

Ref Expression
Assertion csbcnvgALT ( 𝐴𝑉 𝐴 / 𝑥 𝐹 = 𝐴 / 𝑥 𝐹 )

Proof

Step Hyp Ref Expression
1 sbcbr123 ( [ 𝐴 / 𝑥 ] 𝑧 𝐹 𝑦 𝐴 / 𝑥 𝑧 𝐴 / 𝑥 𝐹 𝐴 / 𝑥 𝑦 )
2 csbconstg ( 𝐴𝑉 𝐴 / 𝑥 𝑧 = 𝑧 )
3 csbconstg ( 𝐴𝑉 𝐴 / 𝑥 𝑦 = 𝑦 )
4 2 3 breq12d ( 𝐴𝑉 → ( 𝐴 / 𝑥 𝑧 𝐴 / 𝑥 𝐹 𝐴 / 𝑥 𝑦𝑧 𝐴 / 𝑥 𝐹 𝑦 ) )
5 1 4 syl5bb ( 𝐴𝑉 → ( [ 𝐴 / 𝑥 ] 𝑧 𝐹 𝑦𝑧 𝐴 / 𝑥 𝐹 𝑦 ) )
6 5 opabbidv ( 𝐴𝑉 → { ⟨ 𝑦 , 𝑧 ⟩ ∣ [ 𝐴 / 𝑥 ] 𝑧 𝐹 𝑦 } = { ⟨ 𝑦 , 𝑧 ⟩ ∣ 𝑧 𝐴 / 𝑥 𝐹 𝑦 } )
7 csbopabgALT ( 𝐴𝑉 𝐴 / 𝑥 { ⟨ 𝑦 , 𝑧 ⟩ ∣ 𝑧 𝐹 𝑦 } = { ⟨ 𝑦 , 𝑧 ⟩ ∣ [ 𝐴 / 𝑥 ] 𝑧 𝐹 𝑦 } )
8 df-cnv 𝐴 / 𝑥 𝐹 = { ⟨ 𝑦 , 𝑧 ⟩ ∣ 𝑧 𝐴 / 𝑥 𝐹 𝑦 }
9 8 a1i ( 𝐴𝑉 𝐴 / 𝑥 𝐹 = { ⟨ 𝑦 , 𝑧 ⟩ ∣ 𝑧 𝐴 / 𝑥 𝐹 𝑦 } )
10 6 7 9 3eqtr4rd ( 𝐴𝑉 𝐴 / 𝑥 𝐹 = 𝐴 / 𝑥 { ⟨ 𝑦 , 𝑧 ⟩ ∣ 𝑧 𝐹 𝑦 } )
11 df-cnv 𝐹 = { ⟨ 𝑦 , 𝑧 ⟩ ∣ 𝑧 𝐹 𝑦 }
12 11 csbeq2i 𝐴 / 𝑥 𝐹 = 𝐴 / 𝑥 { ⟨ 𝑦 , 𝑧 ⟩ ∣ 𝑧 𝐹 𝑦 }
13 10 12 eqtr4di ( 𝐴𝑉 𝐴 / 𝑥 𝐹 = 𝐴 / 𝑥 𝐹 )