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Theorem expcomdg 39231
Description: Biconditional form of expcomd 402. (Contributed by Alan Sare, 22-Jul-2012.) (New usage is discouraged.)
Assertion
Ref Expression
expcomdg ((𝜑 → ((𝜓𝜒) → 𝜃)) ↔ (𝜑 → (𝜒 → (𝜓𝜃))))

Proof of Theorem expcomdg
StepHypRef Expression
1 ancomst 454 . . 3 (((𝜓𝜒) → 𝜃) ↔ ((𝜒𝜓) → 𝜃))
2 impexp 437 . . 3 (((𝜒𝜓) → 𝜃) ↔ (𝜒 → (𝜓𝜃)))
31, 2bitri 264 . 2 (((𝜓𝜒) → 𝜃) ↔ (𝜒 → (𝜓𝜃)))
43imbi2i 325 1 ((𝜑 → ((𝜓𝜒) → 𝜃)) ↔ (𝜑 → (𝜒 → (𝜓𝜃))))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 196  wa 382
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8
This theorem depends on definitions:  df-bi 197  df-an 383
This theorem is referenced by: (None)
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