Metamath Proof Explorer


Theorem imadisjlnd

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

Ref Expression
Hypothesis imadisjlnd.1 ⊢ ( 𝜑 → ( dom 𝐴 ∩ 𝐵 ) ≠ ∅ )
Assertion imadisjlnd ( 𝜑 → ( 𝐴 “ 𝐵 ) ≠ ∅ )

Proof

Step Hyp Ref Expression
1 imadisjlnd.1 ⊢ ( 𝜑 → ( dom 𝐴 ∩ 𝐵 ) ≠ ∅ )
2 imadisj ⊢ ( ( 𝐴 “ 𝐵 ) = ∅ ↔ ( dom 𝐴 ∩ 𝐵 ) = ∅ )
3 2 biimpi ⊢ ( ( 𝐴 “ 𝐵 ) = ∅ → ( dom 𝐴 ∩ 𝐵 ) = ∅ )
4 3 necon3i ⊢ ( ( dom 𝐴 ∩ 𝐵 ) ≠ ∅ → ( 𝐴 “ 𝐵 ) ≠ ∅ )
5 1 4 syl ⊢ ( 𝜑 → ( 𝐴 “ 𝐵 ) ≠ ∅ )