Theorem hb3or 706

Description: If x is not free in φ, ψ, and χ, it is not free in (φ ∨ ψ ∨ χ).

Hypotheses

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

Assertion

Ref	Expression
hb3or	⊢ ((φ ∨ ψ ∨ χ) → ∀x(φ ∨ ψ ∨ χ))

Proof of Theorem hb3or

Step	Hyp	Ref	Expression
1		hb.1	. . . 4 ⊢ (φ → ∀xφ)
2		hb.2	. . . 4 ⊢ (ψ → ∀xψ)
3	1, 2	hbor 703	. . 3 ⊢ ((φ ∨ ψ) → ∀x(φ ∨ ψ))
4		hb.3	. . 3 ⊢ (χ → ∀xχ)
5	3, 4	hbor 703	. 2 ⊢ (((φ ∨ ψ) ∨ χ) → ∀x((φ ∨ ψ) ∨ χ))
6		df-3or 582	. 2 ⊢ ((φ ∨ ψ ∨ χ) ↔ ((φ ∨ ψ) ∨ χ))
7	6	bial 695	. 2 ⊢ (∀x(φ ∨ ψ ∨ χ) ↔ ∀x((φ ∨ ψ) ∨ χ))
8	5, 6, 7	3imtr4 192	1 ⊢ ((φ ∨ ψ ∨ χ) → ∀x(φ ∨ ψ ∨ χ))

Syntax hints:   → wi 2   ∨ wo 195   ∨ w3o 580  ∀wal 672

This theorem is referenced by:  hbrdg 2974

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-gen 677

This theorem depends on definitions:  df-bi 128  df-or 197  df-an 198  df-3or 582

metamath.org