Theorem nan 692

Description: Theorem to move a conjunct in and out of a negation. (Contributed by NM, 9-Nov-2003.)

Assertion

Ref	Expression
nan	⊢ ((𝜑 → ¬ (𝜓 ∧ 𝜒)) ↔ ((𝜑 ∧ 𝜓) → ¬ 𝜒))

Proof of Theorem nan

Step	Hyp	Ref	Expression
1		impexp 263	. 2 ⊢ (((𝜑 ∧ 𝜓) → ¬ 𝜒) ↔ (𝜑 → (𝜓 → ¬ 𝜒)))
2		imnan 690	. . 3 ⊢ ((𝜓 → ¬ 𝜒) ↔ ¬ (𝜓 ∧ 𝜒))
3	2	imbi2i 226	. 2 ⊢ ((𝜑 → (𝜓 → ¬ 𝜒)) ↔ (𝜑 → ¬ (𝜓 ∧ 𝜒)))
4	1, 3	bitr2i 185	1 ⊢ ((𝜑 → ¬ (𝜓 ∧ 𝜒)) ↔ ((𝜑 ∧ 𝜓) → ¬ 𝜒))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 104   ↔ wb 105

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 614  ax-in2 615

This theorem depends on definitions:  df-bi 117

This theorem is referenced by:  pm4.15  694