Metamath Proof Explorer


Theorem eldifsnbd

Description: Membership in a set with an element removed implies non-equality with that element. (Contributed by Thierry Arnoux, 13-Jul-2026)

Ref Expression
Hypothesis eldifsnbd.1 ⊢ ( 𝜑 → 𝐴 ∈ ( 𝐵 ∖ { 𝐶 } ) )
Assertion eldifsnbd ( 𝜑 → 𝐴 ≠ 𝐶 )

Proof

Step Hyp Ref Expression
1 eldifsnbd.1 ⊢ ( 𝜑 → 𝐴 ∈ ( 𝐵 ∖ { 𝐶 } ) )
2 eldifsn ⊢ ( 𝐴 ∈ ( 𝐵 ∖ { 𝐶 } ) ↔ ( 𝐴 ∈ 𝐵 ∧ 𝐴 ≠ 𝐶 ) )
3 1 2 sylib ⊢ ( 𝜑 → ( 𝐴 ∈ 𝐵 ∧ 𝐴 ≠ 𝐶 ) )
4 3 simprd ⊢ ( 𝜑 → 𝐴 ≠ 𝐶 )