Metamath Proof Explorer


Theorem imdistand

Description: Distribution of implication with conjunction (deduction form). (Contributed by NM, 27-Aug-2004)

Ref Expression
Hypothesis imdistand.1 ( 𝜑 → ( 𝜓 → ( 𝜒𝜃 ) ) )
Assertion imdistand ( 𝜑 → ( ( 𝜓𝜒 ) → ( 𝜓𝜃 ) ) )

Proof

Step Hyp Ref Expression
1 imdistand.1 ( 𝜑 → ( 𝜓 → ( 𝜒𝜃 ) ) )
2 imdistan ( ( 𝜓 → ( 𝜒𝜃 ) ) ↔ ( ( 𝜓𝜒 ) → ( 𝜓𝜃 ) ) )
3 1 2 sylib ( 𝜑 → ( ( 𝜓𝜒 ) → ( 𝜓𝜃 ) ) )