Theorem nabctnabc 43166

Description: not ( a -> ( b /\ c ) ) we can show: not a implies ( b /\ c ). (Contributed by Jarvin Udandy, 7-Sep-2020.)

Hypothesis

Ref	Expression
nabctnabc.1	⊢ ¬ (𝜑 → (𝜓 ∧ 𝜒))

Assertion

Ref	Expression
nabctnabc	⊢ (¬ 𝜑 → (𝜓 ∧ 𝜒))

Proof of Theorem nabctnabc

Step	Hyp	Ref	Expression
1		nabctnabc.1	. . . . . . . 8 ⊢ ¬ (𝜑 → (𝜓 ∧ 𝜒))
2		pm4.61 407	. . . . . . . . 9 ⊢ (¬ (𝜑 → (𝜓 ∧ 𝜒)) ↔ (𝜑 ∧ ¬ (𝜓 ∧ 𝜒)))
3	2	biimpi 218	. . . . . . . 8 ⊢ (¬ (𝜑 → (𝜓 ∧ 𝜒)) → (𝜑 ∧ ¬ (𝜓 ∧ 𝜒)))
4	1, 3	ax-mp 5	. . . . . . 7 ⊢ (𝜑 ∧ ¬ (𝜓 ∧ 𝜒))
5	4	simpli 486	. . . . . 6 ⊢ 𝜑
6	4	simpri 488	. . . . . 6 ⊢ ¬ (𝜓 ∧ 𝜒)
7	5, 6	2th 266	. . . . 5 ⊢ (𝜑 ↔ ¬ (𝜓 ∧ 𝜒))
8		bicom 224	. . . . . 6 ⊢ ((𝜑 ↔ ¬ (𝜓 ∧ 𝜒)) ↔ (¬ (𝜓 ∧ 𝜒) ↔ 𝜑))
9	8	biimpi 218	. . . . 5 ⊢ ((𝜑 ↔ ¬ (𝜓 ∧ 𝜒)) → (¬ (𝜓 ∧ 𝜒) ↔ 𝜑))
10	7, 9	ax-mp 5	. . . 4 ⊢ (¬ (𝜓 ∧ 𝜒) ↔ 𝜑)
11	10	biimpi 218	. . 3 ⊢ (¬ (𝜓 ∧ 𝜒) → 𝜑)
12	11	con3i 157	. 2 ⊢ (¬ 𝜑 → ¬ ¬ (𝜓 ∧ 𝜒))
13	12	notnotrd 135	1 ⊢ (¬ 𝜑 → (𝜓 ∧ 𝜒))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 208   ∧ wa 398

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

This theorem is referenced by: (None)