Metamath Proof Explorer


Theorem pssnel

Description: A proper subclass has a member in one argument that's not in both. (Contributed by NM, 29-Feb-1996)

Ref Expression
Assertion pssnel ( 𝐴𝐵 → ∃ 𝑥 ( 𝑥𝐵 ∧ ¬ 𝑥𝐴 ) )

Proof

Step Hyp Ref Expression
1 pssdif ( 𝐴𝐵 → ( 𝐵𝐴 ) ≠ ∅ )
2 n0 ( ( 𝐵𝐴 ) ≠ ∅ ↔ ∃ 𝑥 𝑥 ∈ ( 𝐵𝐴 ) )
3 1 2 sylib ( 𝐴𝐵 → ∃ 𝑥 𝑥 ∈ ( 𝐵𝐴 ) )
4 eldif ( 𝑥 ∈ ( 𝐵𝐴 ) ↔ ( 𝑥𝐵 ∧ ¬ 𝑥𝐴 ) )
5 4 exbii ( ∃ 𝑥 𝑥 ∈ ( 𝐵𝐴 ) ↔ ∃ 𝑥 ( 𝑥𝐵 ∧ ¬ 𝑥𝐴 ) )
6 3 5 sylib ( 𝐴𝐵 → ∃ 𝑥 ( 𝑥𝐵 ∧ ¬ 𝑥𝐴 ) )