Theorem hb3an 1482

Description: If 𝑥 is not free in 𝜑, 𝜓, and 𝜒, it is not free in (𝜑 ∧ 𝜓 ∧ 𝜒). (Contributed by NM, 14-Sep-2003.)

Hypotheses

Ref	Expression
hb.1	⊢ (𝜑 → ∀𝑥𝜑)
hb.2	⊢ (𝜓 → ∀𝑥𝜓)
hb.3	⊢ (𝜒 → ∀𝑥𝜒)

Assertion

Ref	Expression
hb3an	⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) → ∀𝑥(𝜑 ∧ 𝜓 ∧ 𝜒))

Proof of Theorem hb3an

Step	Hyp	Ref	Expression
1		df-3an 921	. 2 ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) ↔ ((𝜑 ∧ 𝜓) ∧ 𝜒))
2		hb.1	. . . 4 ⊢ (𝜑 → ∀𝑥𝜑)
3		hb.2	. . . 4 ⊢ (𝜓 → ∀𝑥𝜓)
4	2, 3	hban 1479	. . 3 ⊢ ((𝜑 ∧ 𝜓) → ∀𝑥(𝜑 ∧ 𝜓))
5		hb.3	. . 3 ⊢ (𝜒 → ∀𝑥𝜒)
6	4, 5	hban 1479	. 2 ⊢ (((𝜑 ∧ 𝜓) ∧ 𝜒) → ∀𝑥((𝜑 ∧ 𝜓) ∧ 𝜒))
7	1, 6	hbxfrbi 1401	1 ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) → ∀𝑥(𝜑 ∧ 𝜓 ∧ 𝜒))

Syntax hints:   → wi 4   ∧ wa 102   ∧ w3a 919  ∀wal 1282

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-5 1376  ax-gen 1378

This theorem depends on definitions:  df-bi 115  df-3an 921

This theorem is referenced by: (None)