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Metamath Proof Explorer

Theorem elrabrd

Description: Deduction version of elrab , just like elrabd , but backwards direction. (Contributed by Thierry Arnoux, 15-Jan-2026)

Ref Expression
Hypotheses elrabrd.1 ⊢ ( 𝑥 = 𝐴 → ( 𝜓 ↔ 𝜒 ) )
elrabrd.2 ⊢ ( 𝜑 → 𝐴 ∈ { 𝑥 ∈ 𝐵 ∣ 𝜓 } )
Assertion elrabrd ( 𝜑 → 𝜒 )

Proof

Step Hyp Ref Expression
1 elrabrd.1 ⊢ ( 𝑥 = 𝐴 → ( 𝜓 ↔ 𝜒 ) )
2 elrabrd.2 ⊢ ( 𝜑 → 𝐴 ∈ { 𝑥 ∈ 𝐵 ∣ 𝜓 } )
3 1 elrab ⊢ ( 𝐴 ∈ { 𝑥 ∈ 𝐵 ∣ 𝜓 } ↔ ( 𝐴 ∈ 𝐵 ∧ 𝜒 ) )
4 2 3 sylib ⊢ ( 𝜑 → ( 𝐴 ∈ 𝐵 ∧ 𝜒 ) )
5 4 simprd ⊢ ( 𝜑 → 𝜒 )