Theorem anxordi 1395

Description: Conjunction distributes over exclusive-or. (Contributed by Mario Carneiro and Jim Kingdon, 7-Oct-2018.)

Assertion

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

Proof of Theorem anxordi

Step	Hyp	Ref	Expression
1		simpl 108	. 2 ⊢ ((𝜑 ∧ (𝜓 ⊻ 𝜒)) → 𝜑)
2		df-xor 1371	. . . 4 ⊢ (((𝜑 ∧ 𝜓) ⊻ (𝜑 ∧ 𝜒)) ↔ (((𝜑 ∧ 𝜓) ∨ (𝜑 ∧ 𝜒)) ∧ ¬ ((𝜑 ∧ 𝜓) ∧ (𝜑 ∧ 𝜒))))
3	2	simplbi 272	. . 3 ⊢ (((𝜑 ∧ 𝜓) ⊻ (𝜑 ∧ 𝜒)) → ((𝜑 ∧ 𝜓) ∨ (𝜑 ∧ 𝜒)))
4		simpl 108	. . . 4 ⊢ ((𝜑 ∧ 𝜓) → 𝜑)
5		simpl 108	. . . 4 ⊢ ((𝜑 ∧ 𝜒) → 𝜑)
6	4, 5	jaoi 711	. . 3 ⊢ (((𝜑 ∧ 𝜓) ∨ (𝜑 ∧ 𝜒)) → 𝜑)
7	3, 6	syl 14	. 2 ⊢ (((𝜑 ∧ 𝜓) ⊻ (𝜑 ∧ 𝜒)) → 𝜑)
8		ibar 299	. . 3 ⊢ (𝜑 → ((𝜓 ⊻ 𝜒) ↔ (𝜑 ∧ (𝜓 ⊻ 𝜒))))
9		ibar 299	. . . 4 ⊢ (𝜑 → (𝜓 ↔ (𝜑 ∧ 𝜓)))
10		ibar 299	. . . 4 ⊢ (𝜑 → (𝜒 ↔ (𝜑 ∧ 𝜒)))
11	9, 10	xorbi12d 1377	. . 3 ⊢ (𝜑 → ((𝜓 ⊻ 𝜒) ↔ ((𝜑 ∧ 𝜓) ⊻ (𝜑 ∧ 𝜒))))
12	8, 11	bitr3d 189	. 2 ⊢ (𝜑 → ((𝜑 ∧ (𝜓 ⊻ 𝜒)) ↔ ((𝜑 ∧ 𝜓) ⊻ (𝜑 ∧ 𝜒))))
13	1, 7, 12	pm5.21nii 699	1 ⊢ ((𝜑 ∧ (𝜓 ⊻ 𝜒)) ↔ ((𝜑 ∧ 𝜓) ⊻ (𝜑 ∧ 𝜒)))

Syntax hints:  ¬ wn 3   ∧ wa 103   ↔ wb 104   ∨ wo 703   ⊻ wxo 1370

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-in1 609  ax-in2 610  ax-io 704

This theorem depends on definitions:  df-bi 116  df-xor 1371

This theorem is referenced by:  rpnegap  9643