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Theorem mpjaod 625
Description: Eliminate a disjunction in a deduction. (Contributed by Mario Carneiro, 29-May-2016.)
Hypotheses
Ref Expression
jaod.1 (φ → (ψχ))
jaod.2 (φ → (θχ))
jaod.3 (φ → (ψ θ))
Assertion
Ref Expression
mpjaod (φχ)

Proof of Theorem mpjaod
StepHypRef Expression
1 jaod.3 . 2 (φ → (ψ θ))
2 jaod.1 . . 3 (φ → (ψχ))
3 jaod.2 . . 3 (φ → (θχ))
42, 3jaod 624 . 2 (φ → ((ψ θ) → χ))
51, 4mpd 13 1 (φχ)
Colors of variables: wff set class
Syntax hints:  wi 4   wo 616
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 617
This theorem depends on definitions:  df-bi 110
This theorem is referenced by:  opth1  3947  suctr  4108  onsucelsucexmidlem  4198  reldmtpos  5790  dftpos4  5800  nnm00  6013  enq0tr  6289  prarloclem3step  6350  distrlem4prl  6423  distrlem4pru  6424
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