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Theorem hban 1455
Description: If 𝑥 is not free in 𝜑 and 𝜓, it is not free in (𝜑𝜓). (Contributed by NM, 5-Aug-1993.) (Proof shortened by Mario Carneiro, 2-Feb-2015.)
Hypotheses
Ref Expression
hb.1 (𝜑 → ∀𝑥𝜑)
hb.2 (𝜓 → ∀𝑥𝜓)
Assertion
Ref Expression
hban ((𝜑𝜓) → ∀𝑥(𝜑𝜓))

Proof of Theorem hban
StepHypRef Expression
1 hb.1 . . 3 (𝜑 → ∀𝑥𝜑)
2 hb.2 . . 3 (𝜓 → ∀𝑥𝜓)
31, 2anim12i 325 . 2 ((𝜑𝜓) → (∀𝑥𝜑 ∧ ∀𝑥𝜓))
4 19.26 1386 . 2 (∀𝑥(𝜑𝜓) ↔ (∀𝑥𝜑 ∧ ∀𝑥𝜓))
53, 4sylibr 141 1 ((𝜑𝜓) → ∀𝑥(𝜑𝜓))
Colors of variables: wff set class
Syntax hints:  wi 4  wa 101  wal 1257
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 103  ax-ia2 104  ax-ia3 105  ax-5 1352  ax-gen 1354
This theorem depends on definitions:  df-bi 114
This theorem is referenced by:  hbbi  1456  hb3an  1458  hbsbv  1833  mopick  1994  eupicka  1996  mopick2  1999  cleqh  2153
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