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Theorem nf3or 1906
 Description: If 𝑥 is not free in 𝜑, 𝜓, and 𝜒, then it is not free in (𝜑 ∨ 𝜓 ∨ 𝜒). (Contributed by Mario Carneiro, 11-Aug-2016.)
Hypotheses
Ref Expression
nf.1 𝑥𝜑
nf.2 𝑥𝜓
nf.3 𝑥𝜒
Assertion
Ref Expression
nf3or 𝑥(𝜑𝜓𝜒)

Proof of Theorem nf3or
StepHypRef Expression
1 df-3or 1085 . 2 ((𝜑𝜓𝜒) ↔ ((𝜑𝜓) ∨ 𝜒))
2 nf.1 . . . 4 𝑥𝜑
3 nf.2 . . . 4 𝑥𝜓
42, 3nfor 1905 . . 3 𝑥(𝜑𝜓)
5 nf.3 . . 3 𝑥𝜒
64, 5nfor 1905 . 2 𝑥((𝜑𝜓) ∨ 𝜒)
71, 6nfxfr 1854 1 𝑥(𝜑𝜓𝜒)
 Colors of variables: wff setvar class Syntax hints:   ∨ wo 844   ∨ w3o 1083  Ⅎwnf 1785 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811 This theorem depends on definitions:  df-bi 210  df-an 400  df-or 845  df-3or 1085  df-ex 1782  df-nf 1786 This theorem is referenced by:  nfso  5445
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