Metamath Proof Explorer


Theorem eqelb

Description: Substitution of equal classes into elementhood relation. (Contributed by Peter Mazsa, 17-Jul-2019)

Ref Expression
Assertion eqelb ( ( 𝐴 = 𝐵𝐴𝐶 ) ↔ ( 𝐴 = 𝐵𝐵𝐶 ) )

Proof

Step Hyp Ref Expression
1 simpl ( ( 𝐵 = 𝐴𝐴𝐶 ) → 𝐵 = 𝐴 )
2 eqeltr ( ( 𝐵 = 𝐴𝐴𝐶 ) → 𝐵𝐶 )
3 1 2 jca ( ( 𝐵 = 𝐴𝐴𝐶 ) → ( 𝐵 = 𝐴𝐵𝐶 ) )
4 eqcom ( 𝐵 = 𝐴𝐴 = 𝐵 )
5 4 anbi1i ( ( 𝐵 = 𝐴𝐴𝐶 ) ↔ ( 𝐴 = 𝐵𝐴𝐶 ) )
6 4 anbi1i ( ( 𝐵 = 𝐴𝐵𝐶 ) ↔ ( 𝐴 = 𝐵𝐵𝐶 ) )
7 3 5 6 3imtr3i ( ( 𝐴 = 𝐵𝐴𝐶 ) → ( 𝐴 = 𝐵𝐵𝐶 ) )
8 simpl ( ( 𝐴 = 𝐵𝐵𝐶 ) → 𝐴 = 𝐵 )
9 eqeltr ( ( 𝐴 = 𝐵𝐵𝐶 ) → 𝐴𝐶 )
10 8 9 jca ( ( 𝐴 = 𝐵𝐵𝐶 ) → ( 𝐴 = 𝐵𝐴𝐶 ) )
11 7 10 impbii ( ( 𝐴 = 𝐵𝐴𝐶 ) ↔ ( 𝐴 = 𝐵𝐵𝐶 ) )