Theorem xordidc 1410

Description: Conjunction distributes over exclusive-or, for decidable propositions. This is one way to interpret the distributive law of multiplication over addition in modulo 2 arithmetic. (Contributed by Jim Kingdon, 14-Jul-2018.)

Assertion

Ref	Expression
xordidc	⊢ (DECID 𝜑 → (DECID 𝜓 → (DECID 𝜒 → ((𝜑 ∧ (𝜓 ⊻ 𝜒)) ↔ ((𝜑 ∧ 𝜓) ⊻ (𝜑 ∧ 𝜒))))))

Proof of Theorem xordidc

Step	Hyp	Ref	Expression
1		dcbi 938	. . . . 5 ⊢ (DECID 𝜓 → (DECID 𝜒 → DECID (𝜓 ↔ 𝜒)))
2	1	imp 124	. . . 4 ⊢ ((DECID 𝜓 ∧ DECID 𝜒) → DECID (𝜓 ↔ 𝜒))
3		annimdc 939	. . . . . 6 ⊢ (DECID 𝜑 → (DECID (𝜓 ↔ 𝜒) → ((𝜑 ∧ ¬ (𝜓 ↔ 𝜒)) ↔ ¬ (𝜑 → (𝜓 ↔ 𝜒)))))
4	3	imp 124	. . . . 5 ⊢ ((DECID 𝜑 ∧ DECID (𝜓 ↔ 𝜒)) → ((𝜑 ∧ ¬ (𝜓 ↔ 𝜒)) ↔ ¬ (𝜑 → (𝜓 ↔ 𝜒))))
5		pm5.32 453	. . . . . 6 ⊢ ((𝜑 → (𝜓 ↔ 𝜒)) ↔ ((𝜑 ∧ 𝜓) ↔ (𝜑 ∧ 𝜒)))
6	5	notbii 669	. . . . 5 ⊢ (¬ (𝜑 → (𝜓 ↔ 𝜒)) ↔ ¬ ((𝜑 ∧ 𝜓) ↔ (𝜑 ∧ 𝜒)))
7	4, 6	bitrdi 196	. . . 4 ⊢ ((DECID 𝜑 ∧ DECID (𝜓 ↔ 𝜒)) → ((𝜑 ∧ ¬ (𝜓 ↔ 𝜒)) ↔ ¬ ((𝜑 ∧ 𝜓) ↔ (𝜑 ∧ 𝜒))))
8	2, 7	sylan2 286	. . 3 ⊢ ((DECID 𝜑 ∧ (DECID 𝜓 ∧ DECID 𝜒)) → ((𝜑 ∧ ¬ (𝜓 ↔ 𝜒)) ↔ ¬ ((𝜑 ∧ 𝜓) ↔ (𝜑 ∧ 𝜒))))
9		xornbidc 1402	. . . . . 6 ⊢ (DECID 𝜓 → (DECID 𝜒 → ((𝜓 ⊻ 𝜒) ↔ ¬ (𝜓 ↔ 𝜒))))
10	9	imp 124	. . . . 5 ⊢ ((DECID 𝜓 ∧ DECID 𝜒) → ((𝜓 ⊻ 𝜒) ↔ ¬ (𝜓 ↔ 𝜒)))
11	10	adantl 277	. . . 4 ⊢ ((DECID 𝜑 ∧ (DECID 𝜓 ∧ DECID 𝜒)) → ((𝜓 ⊻ 𝜒) ↔ ¬ (𝜓 ↔ 𝜒)))
12	11	anbi2d 464	. . 3 ⊢ ((DECID 𝜑 ∧ (DECID 𝜓 ∧ DECID 𝜒)) → ((𝜑 ∧ (𝜓 ⊻ 𝜒)) ↔ (𝜑 ∧ ¬ (𝜓 ↔ 𝜒))))
13		dcan 935	. . . . 5 ⊢ ((DECID 𝜑 ∧ DECID 𝜓) → DECID (𝜑 ∧ 𝜓))
14	13	adantrr 479	. . . 4 ⊢ ((DECID 𝜑 ∧ (DECID 𝜓 ∧ DECID 𝜒)) → DECID (𝜑 ∧ 𝜓))
15		dcan 935	. . . . 5 ⊢ ((DECID 𝜑 ∧ DECID 𝜒) → DECID (𝜑 ∧ 𝜒))
16	15	adantrl 478	. . . 4 ⊢ ((DECID 𝜑 ∧ (DECID 𝜓 ∧ DECID 𝜒)) → DECID (𝜑 ∧ 𝜒))
17		xornbidc 1402	. . . 4 ⊢ (DECID (𝜑 ∧ 𝜓) → (DECID (𝜑 ∧ 𝜒) → (((𝜑 ∧ 𝜓) ⊻ (𝜑 ∧ 𝜒)) ↔ ¬ ((𝜑 ∧ 𝜓) ↔ (𝜑 ∧ 𝜒)))))
18	14, 16, 17	sylc 62	. . 3 ⊢ ((DECID 𝜑 ∧ (DECID 𝜓 ∧ DECID 𝜒)) → (((𝜑 ∧ 𝜓) ⊻ (𝜑 ∧ 𝜒)) ↔ ¬ ((𝜑 ∧ 𝜓) ↔ (𝜑 ∧ 𝜒))))
19	8, 12, 18	3bitr4d 220	. 2 ⊢ ((DECID 𝜑 ∧ (DECID 𝜓 ∧ DECID 𝜒)) → ((𝜑 ∧ (𝜓 ⊻ 𝜒)) ↔ ((𝜑 ∧ 𝜓) ⊻ (𝜑 ∧ 𝜒))))
20	19	exp32 365	1 ⊢ (DECID 𝜑 → (DECID 𝜓 → (DECID 𝜒 → ((𝜑 ∧ (𝜓 ⊻ 𝜒)) ↔ ((𝜑 ∧ 𝜓) ⊻ (𝜑 ∧ 𝜒))))))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 104   ↔ wb 105  DECID wdc 835   ⊻ wxo 1386

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 615  ax-in2 616  ax-io 710

This theorem depends on definitions:  df-bi 117  df-stab 832  df-dc 836  df-xor 1387

This theorem is referenced by: (None)