Metamath Proof Explorer


Theorem casesifp

Description: Version of cases expressed using if- . Case disjunction according to the value of ph . One can see this as a proof that the two hypotheses characterize the conditional operator for propositions. For the converses, see ifptru and ifpfal . (Contributed by BJ, 20-Sep-2019)

Ref Expression
Hypotheses casesifp.1
|- ( ph -> ( ps <-> ch ) )
casesifp.2
|- ( -. ph -> ( ps <-> th ) )
Assertion casesifp
|- ( ps <-> if- ( ph , ch , th ) )

Proof

Step Hyp Ref Expression
1 casesifp.1
 |-  ( ph -> ( ps <-> ch ) )
2 casesifp.2
 |-  ( -. ph -> ( ps <-> th ) )
3 1 2 cases
 |-  ( ps <-> ( ( ph /\ ch ) \/ ( -. ph /\ th ) ) )
4 df-ifp
 |-  ( if- ( ph , ch , th ) <-> ( ( ph /\ ch ) \/ ( -. ph /\ th ) ) )
5 3 4 bitr4i
 |-  ( ps <-> if- ( ph , ch , th ) )