Metamath Proof Explorer


Theorem eldifsnd

Description: Membership in a set with an element removed : deduction version. (Contributed by Thierry Arnoux, 4-May-2025)

Ref Expression
Hypotheses eldifsnd.1 ( 𝜑𝐴𝐵 )
eldifsnd.2 ( 𝜑𝐴𝐶 )
Assertion eldifsnd ( 𝜑𝐴 ∈ ( 𝐵 ∖ { 𝐶 } ) )

Proof

Step Hyp Ref Expression
1 eldifsnd.1 ( 𝜑𝐴𝐵 )
2 eldifsnd.2 ( 𝜑𝐴𝐶 )
3 eldifsn ( 𝐴 ∈ ( 𝐵 ∖ { 𝐶 } ) ↔ ( 𝐴𝐵𝐴𝐶 ) )
4 1 2 3 sylanbrc ( 𝜑𝐴 ∈ ( 𝐵 ∖ { 𝐶 } ) )