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 ↔ ¬ 𝑥 ∈ 𝑦 )
Assertion	notzfaus	⊢ ¬ ∃ 𝑦 ∀ 𝑥 ( 𝑥 ∈ 𝑦 ↔ ( 𝑥 ∈ 𝐴 ∧ 𝜑 ) )

Proof

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

Metamath Proof Explorer

Theorem notzfaus

Proof