Theorem nabctnabc 47945

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		pm4.61 410	. . . . . . . . 9 ⊢ (¬ (𝜑 → (𝜓 ∧ 𝜒)) ↔ (𝜑 ∧ ¬ (𝜓 ∧ 𝜒)))
3	2	biimpi 219	. . . . . . . 8 ⊢ (¬ (𝜑 → (𝜓 ∧ 𝜒)) → (𝜑 ∧ ¬ (𝜓 ∧ 𝜒)))
4	1, 3	ax-mp 5	. . . . . . 7 ⊢ (𝜑 ∧ ¬ (𝜓 ∧ 𝜒))
5	4	simpli 489	. . . . . 6 ⊢ 𝜑
6	4	simpri 491	. . . . . 6 ⊢ ¬ (𝜓 ∧ 𝜒)
7	5, 6	2th 267	. . . . 5 ⊢ (𝜑 ↔ ¬ (𝜓 ∧ 𝜒))
8		bicom 225	. . . . . 6 ⊢ ((𝜑 ↔ ¬ (𝜓 ∧ 𝜒)) ↔ (¬ (𝜓 ∧ 𝜒) ↔ 𝜑))
9	8	biimpi 219	. . . . 5 ⊢ ((𝜑 ↔ ¬ (𝜓 ∧ 𝜒)) → (¬ (𝜓 ∧ 𝜒) ↔ 𝜑))
10	7, 9	ax-mp 5	. . . 4 ⊢ (¬ (𝜓 ∧ 𝜒) ↔ 𝜑)
11	10	biimpi 219	. . 3 ⊢ (¬ (𝜓 ∧ 𝜒) → 𝜑)
12	11	con3i 155	. 2 ⊢ (¬ 𝜑 → ¬ ¬ (𝜓 ∧ 𝜒))
13	12	notnotrd 134	1 ⊢ (¬ 𝜑 → (𝜓 ∧ 𝜒))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ↔ wb 209   ∧ wa 401

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This proof depends on definitions:  df-bi 210  df-an 402

This theorem is used by: (None)