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Metamath Proof Explorer

Theorem ineq12d

Description: Equality deduction for intersection of two classes. (Contributed by NM, 24-Jun-2004) (Proof shortened by Andrew Salmon, 26-Jun-2011)

Ref Expression
Hypotheses ineq1d.1 ⊢ ( 𝜑 → 𝐴 = 𝐵 )
ineq12d.2 ⊢ ( 𝜑 → 𝐶 = 𝐷 )
Assertion ineq12d ( 𝜑 → ( 𝐴 ∩ 𝐶 ) = ( 𝐵 ∩ 𝐷 ) )

Proof

Step Hyp Ref Expression
1 ineq1d.1 ⊢ ( 𝜑 → 𝐴 = 𝐵 )
2 ineq12d.2 ⊢ ( 𝜑 → 𝐶 = 𝐷 )
3 ineq12 ⊢ ( ( 𝐴 = 𝐵 ∧ 𝐶 = 𝐷 ) → ( 𝐴 ∩ 𝐶 ) = ( 𝐵 ∩ 𝐷 ) )
4 1 2 3 syl2anc ⊢ ( 𝜑 → ( 𝐴 ∩ 𝐶 ) = ( 𝐵 ∩ 𝐷 ) )