Theorem sepimandl-P4.9d 415

Description: Separate Left Conjunction from Implication.

Hypothesis

Ref	Expression
sepimandl-P4.9d.1	⊢ (𝛾 → ((𝜑 ∧ 𝜓) → 𝜒))

Assertion

Ref	Expression
sepimandl-P4.9d	⊢ (𝛾 → ((𝜑 → 𝜒) ∨ (𝜓 → 𝜒)))

Proof of Theorem sepimandl-P4.9d

Step	Hyp	Ref	Expression
1		rcp-NDASM3of3 197	. . . . . . 7 ⊢ ((𝛾 ∧ 𝜑 ∧ ¬ 𝜒) → ¬ 𝜒)
2		sepimandl-P4.9d.1	. . . . . . . . 9 ⊢ (𝛾 → ((𝜑 ∧ 𝜓) → 𝜒))
3	2	rcp-NDIMP1add2 212	. . . . . . . 8 ⊢ ((𝛾 ∧ 𝜑 ∧ ¬ 𝜒) → ((𝜑 ∧ 𝜓) → 𝜒))
4	3	trnsp-P3.31c 285	. . . . . . 7 ⊢ ((𝛾 ∧ 𝜑 ∧ ¬ 𝜒) → (¬ 𝜒 → ¬ (𝜑 ∧ 𝜓)))
5	1, 4	ndime-P3.6 171	. . . . . 6 ⊢ ((𝛾 ∧ 𝜑 ∧ ¬ 𝜒) → ¬ (𝜑 ∧ 𝜓))
6		rcp-NDASM2of3 196	. . . . . 6 ⊢ ((𝛾 ∧ 𝜑 ∧ ¬ 𝜒) → 𝜑)
7	5, 6	nprofeliml-P4.6a 389	. . . . 5 ⊢ ((𝛾 ∧ 𝜑 ∧ ¬ 𝜒) → ¬ 𝜓)
8	7	rcp-NDIMI3 225	. . . 4 ⊢ ((𝛾 ∧ 𝜑) → (¬ 𝜒 → ¬ 𝜓))
9	8	trnsp-P3.31d 288	. . 3 ⊢ ((𝛾 ∧ 𝜑) → (𝜓 → 𝜒))
10	9	ndoril-P3.10 175	. 2 ⊢ ((𝛾 ∧ 𝜑) → ((𝜑 → 𝜒) ∨ (𝜓 → 𝜒)))
11		rcp-NDASM2of2 194	. . . . 5 ⊢ ((𝛾 ∧ ¬ 𝜑) → ¬ 𝜑)
12	11	axL1-P3.21 252	. . . 4 ⊢ ((𝛾 ∧ ¬ 𝜑) → (¬ 𝜒 → ¬ 𝜑))
13	12	trnsp-P3.31d 288	. . 3 ⊢ ((𝛾 ∧ ¬ 𝜑) → (𝜑 → 𝜒))
14	13	ndorir-P3.11 176	. 2 ⊢ ((𝛾 ∧ ¬ 𝜑) → ((𝜑 → 𝜒) ∨ (𝜓 → 𝜒)))
15		ndexclmid-P3.16.AC 251	. 2 ⊢ (𝛾 → (𝜑 ∨ ¬ 𝜑))
16	10, 14, 15	rcp-NDORE2 235	1 ⊢ (𝛾 → ((𝜑 → 𝜒) ∨ (𝜓 → 𝜒)))

Syntax hints:  ¬ wff-neg 9   → wff-imp 10   ∧ wff-and 132   ∨ wff-or 144   ∧ wff-rcp-AND3 160

This theorem was proved from axioms:  ax-L1 11  ax-L2 12  ax-L3 13  ax-MP 14

This theorem depends on definitions:  df-bi-D2.1 107  df-and-D2.2 133  df-or-D2.3 145  df-true-D2.4 155  df-rcp-AND3 161

This theorem is referenced by:  sepimandl-P4.9d.RC  416  sepimandl-P4.9d.CL  417