Metamath Proof Explorer


Theorem elfilss

Description: An element belongs to a filter iff any element below it does. (Contributed by Stefan O'Rear, 2-Aug-2015)

Ref Expression
Assertion elfilss ( ( 𝐹 ∈ ( Fil ‘ 𝑋 ) ∧ 𝐴𝑋 ) → ( 𝐴𝐹 ↔ ∃ 𝑡𝐹 𝑡𝐴 ) )

Proof

Step Hyp Ref Expression
1 ibar ( 𝐴𝑋 → ( ∃ 𝑡𝐹 𝑡𝐴 ↔ ( 𝐴𝑋 ∧ ∃ 𝑡𝐹 𝑡𝐴 ) ) )
2 1 adantl ( ( 𝐹 ∈ ( Fil ‘ 𝑋 ) ∧ 𝐴𝑋 ) → ( ∃ 𝑡𝐹 𝑡𝐴 ↔ ( 𝐴𝑋 ∧ ∃ 𝑡𝐹 𝑡𝐴 ) ) )
3 filfbas ( 𝐹 ∈ ( Fil ‘ 𝑋 ) → 𝐹 ∈ ( fBas ‘ 𝑋 ) )
4 elfg ( 𝐹 ∈ ( fBas ‘ 𝑋 ) → ( 𝐴 ∈ ( 𝑋 filGen 𝐹 ) ↔ ( 𝐴𝑋 ∧ ∃ 𝑡𝐹 𝑡𝐴 ) ) )
5 3 4 syl ( 𝐹 ∈ ( Fil ‘ 𝑋 ) → ( 𝐴 ∈ ( 𝑋 filGen 𝐹 ) ↔ ( 𝐴𝑋 ∧ ∃ 𝑡𝐹 𝑡𝐴 ) ) )
6 5 adantr ( ( 𝐹 ∈ ( Fil ‘ 𝑋 ) ∧ 𝐴𝑋 ) → ( 𝐴 ∈ ( 𝑋 filGen 𝐹 ) ↔ ( 𝐴𝑋 ∧ ∃ 𝑡𝐹 𝑡𝐴 ) ) )
7 fgfil ( 𝐹 ∈ ( Fil ‘ 𝑋 ) → ( 𝑋 filGen 𝐹 ) = 𝐹 )
8 7 eleq2d ( 𝐹 ∈ ( Fil ‘ 𝑋 ) → ( 𝐴 ∈ ( 𝑋 filGen 𝐹 ) ↔ 𝐴𝐹 ) )
9 8 adantr ( ( 𝐹 ∈ ( Fil ‘ 𝑋 ) ∧ 𝐴𝑋 ) → ( 𝐴 ∈ ( 𝑋 filGen 𝐹 ) ↔ 𝐴𝐹 ) )
10 2 6 9 3bitr2rd ( ( 𝐹 ∈ ( Fil ‘ 𝑋 ) ∧ 𝐴𝑋 ) → ( 𝐴𝐹 ↔ ∃ 𝑡𝐹 𝑡𝐴 ) )