Metamath Proof Explorer


Theorem ifpdfbi

Description: Define the biconditional as conditional logic operator. (Contributed by RP, 20-Apr-2020) (Proof shortened by Wolf Lammen, 30-Apr-2024) (Proof shortened by Garrett Katz, 25-Jun-2026)

Ref Expression
Assertion ifpdfbi
|- ( ( ph <-> ps ) <-> if- ( ph , ps , -. ps ) )

Proof

Step Hyp Ref Expression
1 dfbi3
 |-  ( ( ph <-> ps ) <-> ( ( ph /\ ps ) \/ ( -. ph /\ -. ps ) ) )
2 df-ifp
 |-  ( if- ( ph , ps , -. ps ) <-> ( ( ph /\ ps ) \/ ( -. ph /\ -. ps ) ) )
3 1 2 bitr4i
 |-  ( ( ph <-> ps ) <-> if- ( ph , ps , -. ps ) )