Theorem hban 1561

Description: If 𝑥 is not free in 𝜑 and 𝜓, it is not free in (𝜑 ∧ 𝜓). (Contributed by NM, 5-Aug-1993.) (Proof shortened by Mario Carneiro, 2-Feb-2015.)

Hypotheses

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

Assertion

Ref	Expression
hban	⊢ ((𝜑 ∧ 𝜓) → ∀𝑥(𝜑 ∧ 𝜓))

Proof of Theorem hban

Step	Hyp	Ref	Expression
1		hb.1	. . 3 ⊢ (𝜑 → ∀𝑥𝜑)
2		hb.2	. . 3 ⊢ (𝜓 → ∀𝑥𝜓)
3	1, 2	anim12i 338	. 2 ⊢ ((𝜑 ∧ 𝜓) → (∀𝑥𝜑 ∧ ∀𝑥𝜓))
4		19.26 1495	. 2 ⊢ (∀𝑥(𝜑 ∧ 𝜓) ↔ (∀𝑥𝜑 ∧ ∀𝑥𝜓))
5	3, 4	sylibr 134	1 ⊢ ((𝜑 ∧ 𝜓) → ∀𝑥(𝜑 ∧ 𝜓))

Syntax hints:   → wi 4   ∧ wa 104  ∀wal 1362

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-5 1461  ax-gen 1463

This theorem depends on definitions:  df-bi 117

This theorem is referenced by:  hbbi  1562  hb3an  1564  hbsbv  1960  mopick  2123  eupicka  2125  mopick2  2128  cleqh  2296