Metamath Proof Explorer


Theorem diffib

Description: Case where diffi is a biconditional. (Contributed by Thierry Arnoux, 27-Jun-2024)

Ref Expression
Assertion diffib BFinAFinABFin

Proof

Step Hyp Ref Expression
1 diffi AFinABFin
2 1 adantl BFinAFinABFin
3 difinf ¬AFinBFin¬ABFin
4 3 ancoms BFin¬AFin¬ABFin
5 4 ex BFin¬AFin¬ABFin
6 5 con4d BFinABFinAFin
7 6 imp BFinABFinAFin
8 2 7 impbida BFinAFinABFin