Mathbox for Peter Mazsa < Previous   Next > Nearby theorems Mirrors  >  Home  >  MPE Home  >  Th. List  >   Mathboxes  >  redundpbi1 Structured version   Visualization version   GIF version

Theorem redundpbi1 35931
 Description: Equivalence of redundancy of propositions. (Contributed by Peter Mazsa, 25-Oct-2022.)
Hypothesis
Ref Expression
redundpbi1.1 (𝜑𝜃)
Assertion
Ref Expression
redundpbi1 ( redund (𝜑, 𝜓, 𝜒) ↔ redund (𝜃, 𝜓, 𝜒))

Proof of Theorem redundpbi1
StepHypRef Expression
1 redundpbi1.1 . . . 4 (𝜑𝜃)
21imbi1i 353 . . 3 ((𝜑𝜓) ↔ (𝜃𝜓))
31anbi1i 626 . . . 4 ((𝜑𝜒) ↔ (𝜃𝜒))
43bibi1i 342 . . 3 (((𝜑𝜒) ↔ (𝜓𝜒)) ↔ ((𝜃𝜒) ↔ (𝜓𝜒)))
52, 4anbi12i 629 . 2 (((𝜑𝜓) ∧ ((𝜑𝜒) ↔ (𝜓𝜒))) ↔ ((𝜃𝜓) ∧ ((𝜃𝜒) ↔ (𝜓𝜒))))
6 df-redundp 35925 . 2 ( redund (𝜑, 𝜓, 𝜒) ↔ ((𝜑𝜓) ∧ ((𝜑𝜒) ↔ (𝜓𝜒))))
7 df-redundp 35925 . 2 ( redund (𝜃, 𝜓, 𝜒) ↔ ((𝜃𝜓) ∧ ((𝜃𝜒) ↔ (𝜓𝜒))))
85, 6, 73bitr4i 306 1 ( redund (𝜑, 𝜓, 𝜒) ↔ redund (𝜃, 𝜓, 𝜒))
 Colors of variables: wff setvar class Syntax hints:   → wi 4   ↔ wb 209   ∧ wa 399   redund wredundp 35540 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 210  df-an 400  df-redundp 35925 This theorem is referenced by:  refrelredund3  35937
 Copyright terms: Public domain W3C validator