Metamath Proof Explorer


Theorem notzfaus

Description: In the Separation Scheme zfauscl , we require that y not occur in ph (which can be generalized to "not be free in"). Here we show special cases of A and ph that result in a contradiction if that requirement is not met. (Contributed by NM, 8-Feb-2006) (Proof shortened by BJ, 18-Nov-2023)

Ref Expression
Hypotheses notzfaus.1 𝐴 = { ∅ }
notzfaus.2 ( 𝜑 ↔ ¬ 𝑥𝑦 )
Assertion notzfaus ¬ ∃ 𝑦𝑥 ( 𝑥𝑦 ↔ ( 𝑥𝐴𝜑 ) )

Proof

Step Hyp Ref Expression
1 notzfaus.1 𝐴 = { ∅ }
2 notzfaus.2 ( 𝜑 ↔ ¬ 𝑥𝑦 )
3 0ex ∅ ∈ V
4 3 snnz { ∅ } ≠ ∅
5 1 4 eqnetri 𝐴 ≠ ∅
6 n0 ( 𝐴 ≠ ∅ ↔ ∃ 𝑥 𝑥𝐴 )
7 5 6 mpbi 𝑥 𝑥𝐴
8 pm5.19 ¬ ( 𝑥𝑦 ↔ ¬ 𝑥𝑦 )
9 ibar ( 𝑥𝐴 → ( 𝜑 ↔ ( 𝑥𝐴𝜑 ) ) )
10 9 2 bitr3di ( 𝑥𝐴 → ( ( 𝑥𝐴𝜑 ) ↔ ¬ 𝑥𝑦 ) )
11 10 bibi2d ( 𝑥𝐴 → ( ( 𝑥𝑦 ↔ ( 𝑥𝐴𝜑 ) ) ↔ ( 𝑥𝑦 ↔ ¬ 𝑥𝑦 ) ) )
12 8 11 mtbiri ( 𝑥𝐴 → ¬ ( 𝑥𝑦 ↔ ( 𝑥𝐴𝜑 ) ) )
13 7 12 eximii 𝑥 ¬ ( 𝑥𝑦 ↔ ( 𝑥𝐴𝜑 ) )
14 exnal ( ∃ 𝑥 ¬ ( 𝑥𝑦 ↔ ( 𝑥𝐴𝜑 ) ) ↔ ¬ ∀ 𝑥 ( 𝑥𝑦 ↔ ( 𝑥𝐴𝜑 ) ) )
15 13 14 mpbi ¬ ∀ 𝑥 ( 𝑥𝑦 ↔ ( 𝑥𝐴𝜑 ) )
16 15 nex ¬ ∃ 𝑦𝑥 ( 𝑥𝑦 ↔ ( 𝑥𝐴𝜑 ) )