Metamath Proof Explorer


Theorem ineqri

Description: Inference from membership to intersection. (Contributed by NM, 21-Jun-1993)

Ref Expression
Hypothesis ineqri.1 ( ( 𝑥𝐴𝑥𝐵 ) ↔ 𝑥𝐶 )
Assertion ineqri ( 𝐴𝐵 ) = 𝐶

Proof

Step Hyp Ref Expression
1 ineqri.1 ( ( 𝑥𝐴𝑥𝐵 ) ↔ 𝑥𝐶 )
2 elin ( 𝑥 ∈ ( 𝐴𝐵 ) ↔ ( 𝑥𝐴𝑥𝐵 ) )
3 2 1 bitri ( 𝑥 ∈ ( 𝐴𝐵 ) ↔ 𝑥𝐶 )
4 3 eqriv ( 𝐴𝐵 ) = 𝐶