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Metamath Proof Explorer

Theorem andnand1

Description: Double and in terms of double nand. (Contributed by Anthony Hart, 2-Sep-2011)

Ref Expression
Assertion andnand1 ( ( 𝜑 ∧ 𝜓 ∧ 𝜒 ) ↔ ( ( 𝜑 ⊼ 𝜓 ⊼ 𝜒 ) ⊼ ( 𝜑 ⊼ 𝜓 ⊼ 𝜒 ) ) )

Proof

Step Hyp Ref Expression
1 3anass ⊢ ( ( 𝜑 ∧ 𝜓 ∧ 𝜒 ) ↔ ( 𝜑 ∧ ( 𝜓 ∧ 𝜒 ) ) )
2 pm4.63 ⊢ ( ¬ ( 𝜓 → ¬ 𝜒 ) ↔ ( 𝜓 ∧ 𝜒 ) )
3 2 anbi2i ⊢ ( ( 𝜑 ∧ ¬ ( 𝜓 → ¬ 𝜒 ) ) ↔ ( 𝜑 ∧ ( 𝜓 ∧ 𝜒 ) ) )
4 annim ⊢ ( ( 𝜑 ∧ ¬ ( 𝜓 → ¬ 𝜒 ) ) ↔ ¬ ( 𝜑 → ( 𝜓 → ¬ 𝜒 ) ) )
5 1 3 4 3bitr2i ⊢ ( ( 𝜑 ∧ 𝜓 ∧ 𝜒 ) ↔ ¬ ( 𝜑 → ( 𝜓 → ¬ 𝜒 ) ) )
6 df-3nand ⊢ ( ( 𝜑 ⊼ 𝜓 ⊼ 𝜒 ) ↔ ( 𝜑 → ( 𝜓 → ¬ 𝜒 ) ) )
7 6 notbii ⊢ ( ¬ ( 𝜑 ⊼ 𝜓 ⊼ 𝜒 ) ↔ ¬ ( 𝜑 → ( 𝜓 → ¬ 𝜒 ) ) )
8 nannot ⊢ ( ¬ ( 𝜑 ⊼ 𝜓 ⊼ 𝜒 ) ↔ ( ( 𝜑 ⊼ 𝜓 ⊼ 𝜒 ) ⊼ ( 𝜑 ⊼ 𝜓 ⊼ 𝜒 ) ) )
9 5 7 8 3bitr2i ⊢ ( ( 𝜑 ∧ 𝜓 ∧ 𝜒 ) ↔ ( ( 𝜑 ⊼ 𝜓 ⊼ 𝜒 ) ⊼ ( 𝜑 ⊼ 𝜓 ⊼ 𝜒 ) ) )