Theorem ifpnannanb 39866

Description: Factor conditional logic operator over nand in terms 2 and 3. (Contributed by RP, 21-Apr-2020.)

Assertion

Ref	Expression
ifpnannanb	⊢ (if-(𝜑, (𝜓 ⊼ 𝜒), (𝜃 ⊼ 𝜏)) ↔ (if-(𝜑, 𝜓, 𝜃) ⊼ if-(𝜑, 𝜒, 𝜏)))

Proof of Theorem ifpnannanb

Step	Hyp	Ref	Expression
1		df-nan 1481	. . 3 ⊢ ((𝜓 ⊼ 𝜒) ↔ ¬ (𝜓 ∧ 𝜒))
2		df-nan 1481	. . 3 ⊢ ((𝜃 ⊼ 𝜏) ↔ ¬ (𝜃 ∧ 𝜏))
3		ifpbi23 39831	. . 3 ⊢ ((((𝜓 ⊼ 𝜒) ↔ ¬ (𝜓 ∧ 𝜒)) ∧ ((𝜃 ⊼ 𝜏) ↔ ¬ (𝜃 ∧ 𝜏))) → (if-(𝜑, (𝜓 ⊼ 𝜒), (𝜃 ⊼ 𝜏)) ↔ if-(𝜑, ¬ (𝜓 ∧ 𝜒), ¬ (𝜃 ∧ 𝜏))))
4	1, 2, 3	mp2an 690	. 2 ⊢ (if-(𝜑, (𝜓 ⊼ 𝜒), (𝜃 ⊼ 𝜏)) ↔ if-(𝜑, ¬ (𝜓 ∧ 𝜒), ¬ (𝜃 ∧ 𝜏)))
5		ifpananb 39865	. . . 4 ⊢ (if-(𝜑, (𝜓 ∧ 𝜒), (𝜃 ∧ 𝜏)) ↔ (if-(𝜑, 𝜓, 𝜃) ∧ if-(𝜑, 𝜒, 𝜏)))
6	5	notbii 322	. . 3 ⊢ (¬ if-(𝜑, (𝜓 ∧ 𝜒), (𝜃 ∧ 𝜏)) ↔ ¬ (if-(𝜑, 𝜓, 𝜃) ∧ if-(𝜑, 𝜒, 𝜏)))
7		ifpnotnotb 39838	. . 3 ⊢ (if-(𝜑, ¬ (𝜓 ∧ 𝜒), ¬ (𝜃 ∧ 𝜏)) ↔ ¬ if-(𝜑, (𝜓 ∧ 𝜒), (𝜃 ∧ 𝜏)))
8		df-nan 1481	. . 3 ⊢ ((if-(𝜑, 𝜓, 𝜃) ⊼ if-(𝜑, 𝜒, 𝜏)) ↔ ¬ (if-(𝜑, 𝜓, 𝜃) ∧ if-(𝜑, 𝜒, 𝜏)))
9	6, 7, 8	3bitr4i 305	. 2 ⊢ (if-(𝜑, ¬ (𝜓 ∧ 𝜒), ¬ (𝜃 ∧ 𝜏)) ↔ (if-(𝜑, 𝜓, 𝜃) ⊼ if-(𝜑, 𝜒, 𝜏)))
10	4, 9	bitri 277	1 ⊢ (if-(𝜑, (𝜓 ⊼ 𝜒), (𝜃 ⊼ 𝜏)) ↔ (if-(𝜑, 𝜓, 𝜃) ⊼ if-(𝜑, 𝜒, 𝜏)))

Syntax hints:  ¬ wn 3   ↔ wb 208   ∧ wa 398  if-wif 1057   ⊼ wnan 1480

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-ifp 1058  df-nan 1481

This theorem is referenced by: (None)