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Theorem hb3or 1486
Description: If 𝑥 is not free in 𝜑, 𝜓, and 𝜒, it is not free in (𝜑𝜓𝜒). (Contributed by NM, 14-Sep-2003.)
Hypotheses
Ref Expression
hb.1 (𝜑 → ∀𝑥𝜑)
hb.2 (𝜓 → ∀𝑥𝜓)
hb.3 (𝜒 → ∀𝑥𝜒)
Assertion
Ref Expression
hb3or ((𝜑𝜓𝜒) → ∀𝑥(𝜑𝜓𝜒))

Proof of Theorem hb3or
StepHypRef Expression
1 df-3or 925 . 2 ((𝜑𝜓𝜒) ↔ ((𝜑𝜓) ∨ 𝜒))
2 hb.1 . . . 4 (𝜑 → ∀𝑥𝜑)
3 hb.2 . . . 4 (𝜓 → ∀𝑥𝜓)
42, 3hbor 1483 . . 3 ((𝜑𝜓) → ∀𝑥(𝜑𝜓))
5 hb.3 . . 3 (𝜒 → ∀𝑥𝜒)
64, 5hbor 1483 . 2 (((𝜑𝜓) ∨ 𝜒) → ∀𝑥((𝜑𝜓) ∨ 𝜒))
71, 6hbxfrbi 1406 1 ((𝜑𝜓𝜒) → ∀𝑥(𝜑𝜓𝜒))
Colors of variables: wff set class
Syntax hints:  wi 4  wo 664  w3o 923  wal 1287
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-io 665  ax-5 1381  ax-gen 1383
This theorem depends on definitions:  df-bi 115  df-3or 925
This theorem is referenced by: (None)
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