Theorem anxordi 1555

Description: Conjunction distributes over exclusive-or. In intuitionistic logic this assertion is also true, even though xordi 1034 does not necessarily hold, in part because the usual definition of xor is subtly different in intuitionistic logic. (Contributed by David A. Wheeler, 7-Oct-2018.)

Assertion

Ref	Expression
anxordi	⊢ ((𝜑 ∧ (𝜓 ⊻ 𝜒)) ↔ ((𝜑 ∧ 𝜓) ⊻ (𝜑 ∧ 𝜒)))

Proof of Theorem anxordi

Step	Hyp	Ref	Expression
1		xordi 1034	. 2 ⊢ ((𝜑 ∧ ¬ (𝜓 ↔ 𝜒)) ↔ ¬ ((𝜑 ∧ 𝜓) ↔ (𝜑 ∧ 𝜒)))
2		df-xor 1541	. . 3 ⊢ ((𝜓 ⊻ 𝜒) ↔ ¬ (𝜓 ↔ 𝜒))
3	2	anbi2i 635	. 2 ⊢ ((𝜑 ∧ (𝜓 ⊻ 𝜒)) ↔ (𝜑 ∧ ¬ (𝜓 ↔ 𝜒)))
4		df-xor 1541	. 2 ⊢ (((𝜑 ∧ 𝜓) ⊻ (𝜑 ∧ 𝜒)) ↔ ¬ ((𝜑 ∧ 𝜓) ↔ (𝜑 ∧ 𝜒)))
5	1, 3, 4	3bitr4i 306	1 ⊢ ((𝜑 ∧ (𝜓 ⊻ 𝜒)) ↔ ((𝜑 ∧ 𝜓) ⊻ (𝜑 ∧ 𝜒)))

This proof depends on syntax axioms:  ¬ wn 3   ↔ wb 209   ∧ wa 401   ⊻ wxo 1540

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

This proof depends on definitions:  df-bi 210  df-an 402  df-xor 1541

This theorem is used by: (None)