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Theorem pm5.21nd 868
Description: Eliminate an antecedent implied by each side of a biconditional. (Contributed by NM, 20-Nov-2005.) (Proof shortened by Wolf Lammen, 4-Nov-2013.)
Hypotheses
Ref Expression
pm5.21nd.1 ((φ ψ) → θ)
pm5.21nd.2 ((φ χ) → θ)
pm5.21nd.3 (θ → (ψχ))
Assertion
Ref Expression
pm5.21nd (φ → (ψχ))

Proof of Theorem pm5.21nd
StepHypRef Expression
1 pm5.21nd.1 . . 3 ((φ ψ) → θ)
21ex 423 . 2 (φ → (ψθ))
3 pm5.21nd.2 . . 3 ((φ χ) → θ)
43ex 423 . 2 (φ → (χθ))
5 pm5.21nd.3 . . 3 (θ → (ψχ))
65a1i 10 . 2 (φ → (θ → (ψχ)))
72, 4, 6pm5.21ndd 343 1 (φ → (ψχ))
Colors of variables: wff set class
Syntax hints:  wi 4  wb 176   wa 358
This theorem is referenced by:  ideqg  4880  ideqg2  4881  fvelimab  5410
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 8
This theorem depends on definitions:  df-bi 177  df-an 360
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