Metamath Proof Explorer


Theorem imadisjlnd

Description: Natural deduction form of one negated side of imadisj . (Contributed by Stanislas Polu, 9-Mar-2020)

Ref Expression
Hypothesis imadisjlnd.1
|- ( ph -> ( dom A i^i B ) =/= (/) )
Assertion imadisjlnd
|- ( ph -> ( A " B ) =/= (/) )

Proof

Step Hyp Ref Expression
1 imadisjlnd.1
 |-  ( ph -> ( dom A i^i B ) =/= (/) )
2 imadisj
 |-  ( ( A " B ) = (/) <-> ( dom A i^i B ) = (/) )
3 2 biimpi
 |-  ( ( A " B ) = (/) -> ( dom A i^i B ) = (/) )
4 3 necon3i
 |-  ( ( dom A i^i B ) =/= (/) -> ( A " B ) =/= (/) )
5 1 4 syl
 |-  ( ph -> ( A " B ) =/= (/) )