Theorem anxordi 1288

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 102	. 2 ⊢ ((φ ∧ (ψ ⊻ χ)) → φ)
2		df-xor 1266	. . . 4 ⊢ (((φ ∧ ψ) ⊻ (φ ∧ χ)) ↔ (((φ ∧ ψ) ∨ (φ ∧ χ)) ∧ ¬ ((φ ∧ ψ) ∧ (φ ∧ χ))))
3	2	simplbi 259	. . 3 ⊢ (((φ ∧ ψ) ⊻ (φ ∧ χ)) → ((φ ∧ ψ) ∨ (φ ∧ χ)))
4		simpl 102	. . . 4 ⊢ ((φ ∧ ψ) → φ)
5		simpl 102	. . . 4 ⊢ ((φ ∧ χ) → φ)
6	4, 5	jaoi 635	. . 3 ⊢ (((φ ∧ ψ) ∨ (φ ∧ χ)) → φ)
7	3, 6	syl 14	. 2 ⊢ (((φ ∧ ψ) ⊻ (φ ∧ χ)) → φ)
8		ibar 285	. . 3 ⊢ (φ → ((ψ ⊻ χ) ↔ (φ ∧ (ψ ⊻ χ))))
9		ibar 285	. . . 4 ⊢ (φ → (ψ ↔ (φ ∧ ψ)))
10		ibar 285	. . . 4 ⊢ (φ → (χ ↔ (φ ∧ χ)))
11	9, 10	xorbi12d 1271	. . 3 ⊢ (φ → ((ψ ⊻ χ) ↔ ((φ ∧ ψ) ⊻ (φ ∧ χ))))
12	8, 11	bitr3d 179	. 2 ⊢ (φ → ((φ ∧ (ψ ⊻ χ)) ↔ ((φ ∧ ψ) ⊻ (φ ∧ χ))))
13	1, 7, 12	pm5.21nii 619	1 ⊢ ((φ ∧ (ψ ⊻ χ)) ↔ ((φ ∧ ψ) ⊻ (φ ∧ χ)))

Syntax hints:  ¬ wn 3   ∧ wa 97   ↔ wb 98   ∨ wo 628   ⊻ wxo 1265

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-in1 544  ax-in2 545  ax-io 629

This theorem depends on definitions:  df-bi 110  df-xor 1266

This theorem is referenced by:  rpnegap  8390