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Metamath Proof Explorer

Theorem spcdv

Description: Rule of specialization, using implicit substitution. Analogous to rspcdv . (Contributed by David Moews, 1-May-2017)

Ref Expression
Hypotheses spcimdv.1 ⊢ ( 𝜑 → 𝐴 ∈ 𝐵 )
spcdv.2 ⊢ ( ( 𝜑 ∧ 𝑥 = 𝐴 ) → ( 𝜓 ↔ 𝜒 ) )
Assertion spcdv ( 𝜑 → ( ∀ 𝑥 𝜓 → 𝜒 ) )

Proof

Step Hyp Ref Expression
1 spcimdv.1 ⊢ ( 𝜑 → 𝐴 ∈ 𝐵 )
2 spcdv.2 ⊢ ( ( 𝜑 ∧ 𝑥 = 𝐴 ) → ( 𝜓 ↔ 𝜒 ) )
3 2 biimpd ⊢ ( ( 𝜑 ∧ 𝑥 = 𝐴 ) → ( 𝜓 → 𝜒 ) )
4 1 3 spcimdv ⊢ ( 𝜑 → ( ∀ 𝑥 𝜓 → 𝜒 ) )