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Theorem nanbi1 1452
 Description: Introduce a right anti-conjunct to both sides of a logical equivalence. (Contributed by SF, 2-Jan-2018.)
Assertion
Ref Expression
nanbi1 ((𝜑𝜓) → ((𝜑𝜒) ↔ (𝜓𝜒)))

Proof of Theorem nanbi1
StepHypRef Expression
1 anbi1 742 . . 3 ((𝜑𝜓) → ((𝜑𝜒) ↔ (𝜓𝜒)))
21notbid 308 . 2 ((𝜑𝜓) → (¬ (𝜑𝜒) ↔ ¬ (𝜓𝜒)))
3 df-nan 1445 . 2 ((𝜑𝜒) ↔ ¬ (𝜑𝜒))
4 df-nan 1445 . 2 ((𝜓𝜒) ↔ ¬ (𝜓𝜒))
52, 3, 43bitr4g 303 1 ((𝜑𝜓) → ((𝜑𝜒) ↔ (𝜓𝜒)))
 Colors of variables: wff setvar class Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 196   ∧ wa 384   ⊼ wnan 1444 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 386  df-nan 1445 This theorem is referenced by:  nanbi2  1453  nanbi12  1454  nanbi1i  1455  nanbi1d  1458  nabi1  32052
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