Theorem eliminable-abeqv 37701

Description: A theorem used to prove the base case of the Eliminability Theorem (see section comment): abstraction equals variable. (Contributed by BJ, 30-Apr-2024.) Beware not to use symmetry of class equality. (Proof modification is discouraged.) (New usage is discouraged.)

Assertion

Ref	Expression
eliminable-abeqv	⊢ ({𝑥 ∣ 𝜑} = 𝑦 ↔ ∀𝑧([𝑧 / 𝑥]𝜑 ↔ 𝑧 ∈ 𝑦))

Distinct variable groups:   𝑥,𝑧   𝑦,𝑧   𝜑,𝑧

Allowed substitution hints:   𝜑(𝑥, 𝑦)

Proof of Theorem eliminable-abeqv

Step	Hyp	Ref	Expression
1		dfcleq 2753	. 2 ⊢ ({𝑥 ∣ 𝜑} = 𝑦 ↔ ∀𝑧(𝑧 ∈ {𝑥 ∣ 𝜑} ↔ 𝑧 ∈ 𝑦))
2		eliminable-velab 37699	. . . 4 ⊢ (𝑧 ∈ {𝑥 ∣ 𝜑} ↔ [𝑧 / 𝑥]𝜑)
3	2	bibi1i 341	. . 3 ⊢ ((𝑧 ∈ {𝑥 ∣ 𝜑} ↔ 𝑧 ∈ 𝑦) ↔ ([𝑧 / 𝑥]𝜑 ↔ 𝑧 ∈ 𝑦))
4	3	albii 1852	. 2 ⊢ (∀𝑧(𝑧 ∈ {𝑥 ∣ 𝜑} ↔ 𝑧 ∈ 𝑦) ↔ ∀𝑧([𝑧 / 𝑥]𝜑 ↔ 𝑧 ∈ 𝑦))
5	1, 4	bitri 278	1 ⊢ ({𝑥 ∣ 𝜑} = 𝑦 ↔ ∀𝑧([𝑧 / 𝑥]𝜑 ↔ 𝑧 ∈ 𝑦))

This proof depends on syntax axioms:   ↔ wb 209  ∀wal 1568   = wceq 1570  [wsb 2099   ∈ wcel 2145  {cab 2738

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-9 2155  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-clab 2739  df-cleq 2752

This theorem is used by: (None)