Theorem 19.26-3an 1471

Description: Theorem 19.26 of [Margaris] p. 90 with triple conjunction. (Contributed by NM, 13-Sep-2011.)

Assertion

Ref	Expression
19.26-3an	⊢ (∀𝑥(𝜑 ∧ 𝜓 ∧ 𝜒) ↔ (∀𝑥𝜑 ∧ ∀𝑥𝜓 ∧ ∀𝑥𝜒))

Proof of Theorem 19.26-3an

Step	Hyp	Ref	Expression
1		19.26 1469	. . 3 ⊢ (∀𝑥((𝜑 ∧ 𝜓) ∧ 𝜒) ↔ (∀𝑥(𝜑 ∧ 𝜓) ∧ ∀𝑥𝜒))
2		19.26 1469	. . . 4 ⊢ (∀𝑥(𝜑 ∧ 𝜓) ↔ (∀𝑥𝜑 ∧ ∀𝑥𝜓))
3	2	anbi1i 454	. . 3 ⊢ ((∀𝑥(𝜑 ∧ 𝜓) ∧ ∀𝑥𝜒) ↔ ((∀𝑥𝜑 ∧ ∀𝑥𝜓) ∧ ∀𝑥𝜒))
4	1, 3	bitri 183	. 2 ⊢ (∀𝑥((𝜑 ∧ 𝜓) ∧ 𝜒) ↔ ((∀𝑥𝜑 ∧ ∀𝑥𝜓) ∧ ∀𝑥𝜒))
5		df-3an 970	. . 3 ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) ↔ ((𝜑 ∧ 𝜓) ∧ 𝜒))
6	5	albii 1458	. 2 ⊢ (∀𝑥(𝜑 ∧ 𝜓 ∧ 𝜒) ↔ ∀𝑥((𝜑 ∧ 𝜓) ∧ 𝜒))
7		df-3an 970	. 2 ⊢ ((∀𝑥𝜑 ∧ ∀𝑥𝜓 ∧ ∀𝑥𝜒) ↔ ((∀𝑥𝜑 ∧ ∀𝑥𝜓) ∧ ∀𝑥𝜒))
8	4, 6, 7	3bitr4i 211	1 ⊢ (∀𝑥(𝜑 ∧ 𝜓 ∧ 𝜒) ↔ (∀𝑥𝜑 ∧ ∀𝑥𝜓 ∧ ∀𝑥𝜒))

Syntax hints:   ∧ wa 103   ↔ wb 104   ∧ w3a 968  ∀wal 1341

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-5 1435  ax-gen 1437

This theorem depends on definitions:  df-bi 116  df-3an 970

This theorem is referenced by:  hb3and  1478