Theorem hbim 1525

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 O'Cat, 3-Mar-2008.) (Revised by Mario Carneiro, 2-Feb-2015.)

Hypotheses

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

Assertion

Ref	Expression
hbim	⊢ ((𝜑 → 𝜓) → ∀𝑥(𝜑 → 𝜓))

Proof of Theorem hbim

Step	Hyp	Ref	Expression
1		ax-4 1488	. . 3 ⊢ (∀𝑥𝜑 → 𝜑)
2		hb.2	. . 3 ⊢ (𝜓 → ∀𝑥𝜓)
3	1, 2	imim12i 59	. 2 ⊢ ((𝜑 → 𝜓) → (∀𝑥𝜑 → ∀𝑥𝜓))
4		ax-i5r 1516	. 2 ⊢ ((∀𝑥𝜑 → ∀𝑥𝜓) → ∀𝑥(∀𝑥𝜑 → 𝜓))
5		hb.1	. . . 4 ⊢ (𝜑 → ∀𝑥𝜑)
6	5	imim1i 60	. . 3 ⊢ ((∀𝑥𝜑 → 𝜓) → (𝜑 → 𝜓))
7	6	alimi 1432	. 2 ⊢ (∀𝑥(∀𝑥𝜑 → 𝜓) → ∀𝑥(𝜑 → 𝜓))
8	3, 4, 7	3syl 17	1 ⊢ ((𝜑 → 𝜓) → ∀𝑥(𝜑 → 𝜓))

Syntax hints:   → wi 4  ∀wal 1330

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-5 1424  ax-gen 1426  ax-4 1488  ax-i5r 1516

This theorem is referenced by:  hbbi  1528  hbia1  1532  19.21h  1537  19.38  1655  hbsbv  1915  hbmo1  2038  hbmo  2039  moexexdc  2084  2eu4  2093  cleqh  2240  hbral  2467