Metamath Proof Explorer


Theorem difeqri

Description: Inference from membership to difference. (Contributed by NM, 17-May-1998) (Proof shortened by Andrew Salmon, 26-Jun-2011)

Ref Expression
Hypothesis difeqri.1 ( ( 𝑥𝐴 ∧ ¬ 𝑥𝐵 ) ↔ 𝑥𝐶 )
Assertion difeqri ( 𝐴𝐵 ) = 𝐶

Proof

Step Hyp Ref Expression
1 difeqri.1 ( ( 𝑥𝐴 ∧ ¬ 𝑥𝐵 ) ↔ 𝑥𝐶 )
2 eldif ( 𝑥 ∈ ( 𝐴𝐵 ) ↔ ( 𝑥𝐴 ∧ ¬ 𝑥𝐵 ) )
3 2 1 bitri ( 𝑥 ∈ ( 𝐴𝐵 ) ↔ 𝑥𝐶 )
4 3 eqriv ( 𝐴𝐵 ) = 𝐶