Theorem hbim 1419

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 1381	. . 3 ⊢ (∀xφ → φ)
2		hb.2	. . 3 ⊢ (ψ → ∀xψ)
3	1, 2	imim12i 53	. 2 ⊢ ((φ → ψ) → (∀xφ → ∀xψ))
4		ax-i5r 1410	. 2 ⊢ ((∀xφ → ∀xψ) → ∀x(∀xφ → ψ))
5		hb.1	. . . 4 ⊢ (φ → ∀xφ)
6	5	imim1i 54	. . 3 ⊢ ((∀xφ → ψ) → (φ → ψ))
7	6	alimi 1324	. 2 ⊢ (∀x(∀xφ → ψ) → ∀x(φ → ψ))
8	3, 4, 7	3syl 17	1 ⊢ ((φ → ψ) → ∀x(φ → ψ))

Syntax hints:   → wi 4  ∀wal 1226

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-5 1316  ax-gen 1318  ax-4 1381  ax-i5r 1410

This theorem is referenced by:  hbbi  1422  hbia1  1426  19.21h  1431  19.38  1548  hbsbv  1799  hbmo1  1920  hbmo  1921  moexexdc  1966  2eu4  1975  cleqh  2119  hbral  2331