Theorem imdistand 342

Description: Distribution of implication with conjunction (deduction rule).

Hypothesis

Ref	Expression
imdistand.1	⊢ (φ → (ψ → (χ → θ)))

Assertion

Ref	Expression
imdistand	⊢ (φ → ((ψ ∧ χ) → (ψ ∧ θ)))

Proof of Theorem imdistand

Step	Hyp	Ref	Expression
1		imdistand.1	. 2 ⊢ (φ → (ψ → (χ → θ)))
2		imdistan 339	. 2 ⊢ ((ψ → (χ → θ)) ↔ ((ψ ∧ χ) → (ψ ∧ θ)))
3	1, 2	sylib 173	1 ⊢ (φ → ((ψ ∧ χ) → (ψ ∧ θ)))

Syntax hints:   → wi 2   ∧ wa 196

This theorem is referenced by:  fconstfv 2903  zornlem7 3609  cfub 3703  cflim 3704  prlem936b 3948  suppsr3 4018  supsrlem2 4020  ocsh 5164

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6

This theorem depends on definitions:  df-bi 128  df-an 198

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