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Theorem hbor 1412
 Description: If x is not free in φ and ψ, it is not free in (φ ∨ ψ). (Contributed by NM, 5-Aug-1993.) (Revised by NM, 2-Feb-2015.)
Hypotheses
Ref Expression
hb.1 (φxφ)
hb.2 (ψxψ)
Assertion
Ref Expression
hbor ((φ ψ) → x(φ ψ))

Proof of Theorem hbor
StepHypRef Expression
1 hb.1 . . 3 (φxφ)
2 orc 617 . . . 4 (φ → (φ ψ))
32alimi 1318 . . 3 (xφx(φ ψ))
41, 3syl 14 . 2 (φx(φ ψ))
5 hb.2 . . 3 (ψxψ)
6 olc 616 . . . 4 (ψ → (φ ψ))
76alimi 1318 . . 3 (xψx(φ ψ))
85, 7syl 14 . 2 (ψx(φ ψ))
94, 8jaoi 620 1 ((φ ψ) → x(φ ψ))
 Colors of variables: wff set class Syntax hints:   → wi 4   ∨ wo 613  ∀wal 1222 This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 614  ax-5 1310  ax-gen 1312 This theorem depends on definitions:  df-bi 110 This theorem is referenced by:  hb3or  1415  nfor  1440  19.43  1493
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