Metamath Proof Explorer


Theorem rexdifsn

Description: Restricted existential quantification over a set with an element removed. (Contributed by NM, 4-Feb-2015)

Ref Expression
Assertion rexdifsn ( ∃ 𝑥 ∈ ( 𝐴 ∖ { 𝐵 } ) 𝜑 ↔ ∃ 𝑥 ∈ 𝐴 ( 𝑥 ≠ 𝐵 ∧ 𝜑 ) )

Proof

Step Hyp Ref Expression
1 eldifsn ⊢ ( 𝑥 ∈ ( 𝐴 ∖ { 𝐵 } ) ↔ ( 𝑥 ∈ 𝐴 ∧ 𝑥 ≠ 𝐵 ) )
2 1 anbi1i ⊢ ( ( 𝑥 ∈ ( 𝐴 ∖ { 𝐵 } ) ∧ 𝜑 ) ↔ ( ( 𝑥 ∈ 𝐴 ∧ 𝑥 ≠ 𝐵 ) ∧ 𝜑 ) )
3 anass ⊢ ( ( ( 𝑥 ∈ 𝐴 ∧ 𝑥 ≠ 𝐵 ) ∧ 𝜑 ) ↔ ( 𝑥 ∈ 𝐴 ∧ ( 𝑥 ≠ 𝐵 ∧ 𝜑 ) ) )
4 2 3 bitri ⊢ ( ( 𝑥 ∈ ( 𝐴 ∖ { 𝐵 } ) ∧ 𝜑 ) ↔ ( 𝑥 ∈ 𝐴 ∧ ( 𝑥 ≠ 𝐵 ∧ 𝜑 ) ) )
5 4 rexbii2 ⊢ ( ∃ 𝑥 ∈ ( 𝐴 ∖ { 𝐵 } ) 𝜑 ↔ ∃ 𝑥 ∈ 𝐴 ( 𝑥 ≠ 𝐵 ∧ 𝜑 ) )