Theorem abciffcbatnabciffncbai 47999

Description: Operands in a biconditional expression converted negated. Additionally biconditional converted to show antecedent implies sequent. (Contributed by Jarvin Udandy, 7-Sep-2020.)

Hypothesis

Ref	Expression
abciffcbatnabciffncbai.1	⊢ (((𝜑 ∧ 𝜓) ∧ 𝜒) ↔ ((𝜒 ∧ 𝜓) ∧ 𝜑))

Assertion

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

Proof of Theorem abciffcbatnabciffncbai

Step	Hyp	Ref	Expression
1		abciffcbatnabciffncbai.1	. . 3 ⊢ (((𝜑 ∧ 𝜓) ∧ 𝜒) ↔ ((𝜒 ∧ 𝜓) ∧ 𝜑))
2		notbi 322	. . . 4 ⊢ ((((𝜑 ∧ 𝜓) ∧ 𝜒) ↔ ((𝜒 ∧ 𝜓) ∧ 𝜑)) ↔ (¬ ((𝜑 ∧ 𝜓) ∧ 𝜒) ↔ ¬ ((𝜒 ∧ 𝜓) ∧ 𝜑)))
3	2	biimpi 219	. . 3 ⊢ ((((𝜑 ∧ 𝜓) ∧ 𝜒) ↔ ((𝜒 ∧ 𝜓) ∧ 𝜑)) → (¬ ((𝜑 ∧ 𝜓) ∧ 𝜒) ↔ ¬ ((𝜒 ∧ 𝜓) ∧ 𝜑)))
4	1, 3	ax-mp 5	. 2 ⊢ (¬ ((𝜑 ∧ 𝜓) ∧ 𝜒) ↔ ¬ ((𝜒 ∧ 𝜓) ∧ 𝜑))
5	4	biimpi 219	1 ⊢ (¬ ((𝜑 ∧ 𝜓) ∧ 𝜒) → ¬ ((𝜒 ∧ 𝜓) ∧ 𝜑))

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

This theorem is used by: (None)