Theorem xordi 865

Description: Conjunction distributes over exclusive-or, using ¬ (φ ↔ ψ) to express exclusive-or. This is one way to interpret the distributive law of multiplication over addition in modulo 2 arithmetic. (Contributed by NM, 3-Oct-2008.)

Assertion

Ref	Expression
xordi	⊢ ((φ ∧ ¬ (ψ ↔ χ)) ↔ ¬ ((φ ∧ ψ) ↔ (φ ∧ χ)))

Proof of Theorem xordi

Step	Hyp	Ref	Expression
1		annim 414	. 2 ⊢ ((φ ∧ ¬ (ψ ↔ χ)) ↔ ¬ (φ → (ψ ↔ χ)))
2		pm5.32 617	. 2 ⊢ ((φ → (ψ ↔ χ)) ↔ ((φ ∧ ψ) ↔ (φ ∧ χ)))
3	1, 2	xchbinx 301	1 ⊢ ((φ ∧ ¬ (ψ ↔ χ)) ↔ ¬ ((φ ∧ ψ) ↔ (φ ∧ χ)))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ↔ wb 176   ∧ wa 358

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This proof depends on definitions:  df-bi 177  df-an 360

This theorem is used by: (None)