Metamath Proof Explorer


Theorem ab0w

Description: The class of sets verifying a property is the empty class if and only if that property is a contradiction. Version of ab0 using implicit substitution, which requires fewer axioms. (Contributed by Gino Giotto, 3-Oct-2024)

Ref Expression
Hypothesis ab0w.1 ( 𝑥 = 𝑦 → ( 𝜑𝜓 ) )
Assertion ab0w ( { 𝑥𝜑 } = ∅ ↔ ∀ 𝑦 ¬ 𝜓 )

Proof

Step Hyp Ref Expression
1 ab0w.1 ( 𝑥 = 𝑦 → ( 𝜑𝜓 ) )
2 dfnul4 ∅ = { 𝑥 ∣ ⊥ }
3 2 eqeq2i ( { 𝑥𝜑 } = ∅ ↔ { 𝑥𝜑 } = { 𝑥 ∣ ⊥ } )
4 dfcleq ( { 𝑥𝜑 } = { 𝑥 ∣ ⊥ } ↔ ∀ 𝑦 ( 𝑦 ∈ { 𝑥𝜑 } ↔ 𝑦 ∈ { 𝑥 ∣ ⊥ } ) )
5 df-clab ( 𝑦 ∈ { 𝑥𝜑 } ↔ [ 𝑦 / 𝑥 ] 𝜑 )
6 1 sbievw ( [ 𝑦 / 𝑥 ] 𝜑𝜓 )
7 5 6 bitri ( 𝑦 ∈ { 𝑥𝜑 } ↔ 𝜓 )
8 7 bibi1i ( ( 𝑦 ∈ { 𝑥𝜑 } ↔ 𝑦 ∈ { 𝑥 ∣ ⊥ } ) ↔ ( 𝜓𝑦 ∈ { 𝑥 ∣ ⊥ } ) )
9 bicom ( ( 𝜓𝑦 ∈ { 𝑥 ∣ ⊥ } ) ↔ ( 𝑦 ∈ { 𝑥 ∣ ⊥ } ↔ 𝜓 ) )
10 8 9 bitri ( ( 𝑦 ∈ { 𝑥𝜑 } ↔ 𝑦 ∈ { 𝑥 ∣ ⊥ } ) ↔ ( 𝑦 ∈ { 𝑥 ∣ ⊥ } ↔ 𝜓 ) )
11 10 albii ( ∀ 𝑦 ( 𝑦 ∈ { 𝑥𝜑 } ↔ 𝑦 ∈ { 𝑥 ∣ ⊥ } ) ↔ ∀ 𝑦 ( 𝑦 ∈ { 𝑥 ∣ ⊥ } ↔ 𝜓 ) )
12 4 11 bitri ( { 𝑥𝜑 } = { 𝑥 ∣ ⊥ } ↔ ∀ 𝑦 ( 𝑦 ∈ { 𝑥 ∣ ⊥ } ↔ 𝜓 ) )
13 df-clab ( 𝑦 ∈ { 𝑥 ∣ ⊥ } ↔ [ 𝑦 / 𝑥 ] ⊥ )
14 sbv ( [ 𝑦 / 𝑥 ] ⊥ ↔ ⊥ )
15 13 14 bitri ( 𝑦 ∈ { 𝑥 ∣ ⊥ } ↔ ⊥ )
16 15 bibi1i ( ( 𝑦 ∈ { 𝑥 ∣ ⊥ } ↔ 𝜓 ) ↔ ( ⊥ ↔ 𝜓 ) )
17 16 albii ( ∀ 𝑦 ( 𝑦 ∈ { 𝑥 ∣ ⊥ } ↔ 𝜓 ) ↔ ∀ 𝑦 ( ⊥ ↔ 𝜓 ) )
18 falim ( ⊥ → ( 𝜓 → ¬ 𝜓 ) )
19 idd ( ¬ ⊥ → ( ¬ 𝜓 → ¬ 𝜓 ) )
20 18 19 bija ( ( ⊥ ↔ 𝜓 ) → ¬ 𝜓 )
21 falim ( ⊥ → 𝜓 )
22 id ( 𝜓𝜓 )
23 21 22 pm5.21ni ( ¬ 𝜓 → ( ⊥ ↔ 𝜓 ) )
24 20 23 impbii ( ( ⊥ ↔ 𝜓 ) ↔ ¬ 𝜓 )
25 24 albii ( ∀ 𝑦 ( ⊥ ↔ 𝜓 ) ↔ ∀ 𝑦 ¬ 𝜓 )
26 17 25 bitri ( ∀ 𝑦 ( 𝑦 ∈ { 𝑥 ∣ ⊥ } ↔ 𝜓 ) ↔ ∀ 𝑦 ¬ 𝜓 )
27 12 26 bitri ( { 𝑥𝜑 } = { 𝑥 ∣ ⊥ } ↔ ∀ 𝑦 ¬ 𝜓 )
28 3 27 bitri ( { 𝑥𝜑 } = ∅ ↔ ∀ 𝑦 ¬ 𝜓 )