Metamath Proof Explorer


Theorem axpweq

Description: Two equivalent ways to express the Power Set Axiom. Note that ax-pow is not used by the proof. When ax-pow is assumed and A is a set, both sides of the biconditional hold. In ZF, both sides hold if and only if A is a set (see pwexr ). (Contributed by NM, 22-Jun-2009)

Ref Expression
Assertion axpweq ( 𝒫 𝐴 ∈ V ↔ ∃ 𝑥𝑦 ( ∀ 𝑧 ( 𝑧𝑦𝑧𝐴 ) → 𝑦𝑥 ) )

Proof

Step Hyp Ref Expression
1 pwidg ( 𝒫 𝐴 ∈ V → 𝒫 𝐴 ∈ 𝒫 𝒫 𝐴 )
2 pweq ( 𝑥 = 𝒫 𝐴 → 𝒫 𝑥 = 𝒫 𝒫 𝐴 )
3 2 eleq2d ( 𝑥 = 𝒫 𝐴 → ( 𝒫 𝐴 ∈ 𝒫 𝑥 ↔ 𝒫 𝐴 ∈ 𝒫 𝒫 𝐴 ) )
4 3 spcegv ( 𝒫 𝐴 ∈ V → ( 𝒫 𝐴 ∈ 𝒫 𝒫 𝐴 → ∃ 𝑥 𝒫 𝐴 ∈ 𝒫 𝑥 ) )
5 1 4 mpd ( 𝒫 𝐴 ∈ V → ∃ 𝑥 𝒫 𝐴 ∈ 𝒫 𝑥 )
6 elex ( 𝒫 𝐴 ∈ 𝒫 𝑥 → 𝒫 𝐴 ∈ V )
7 6 exlimiv ( ∃ 𝑥 𝒫 𝐴 ∈ 𝒫 𝑥 → 𝒫 𝐴 ∈ V )
8 5 7 impbii ( 𝒫 𝐴 ∈ V ↔ ∃ 𝑥 𝒫 𝐴 ∈ 𝒫 𝑥 )
9 vex 𝑥 ∈ V
10 9 elpw2 ( 𝒫 𝐴 ∈ 𝒫 𝑥 ↔ 𝒫 𝐴𝑥 )
11 pwss ( 𝒫 𝐴𝑥 ↔ ∀ 𝑦 ( 𝑦𝐴𝑦𝑥 ) )
12 dfss2 ( 𝑦𝐴 ↔ ∀ 𝑧 ( 𝑧𝑦𝑧𝐴 ) )
13 12 imbi1i ( ( 𝑦𝐴𝑦𝑥 ) ↔ ( ∀ 𝑧 ( 𝑧𝑦𝑧𝐴 ) → 𝑦𝑥 ) )
14 13 albii ( ∀ 𝑦 ( 𝑦𝐴𝑦𝑥 ) ↔ ∀ 𝑦 ( ∀ 𝑧 ( 𝑧𝑦𝑧𝐴 ) → 𝑦𝑥 ) )
15 11 14 bitri ( 𝒫 𝐴𝑥 ↔ ∀ 𝑦 ( ∀ 𝑧 ( 𝑧𝑦𝑧𝐴 ) → 𝑦𝑥 ) )
16 10 15 bitri ( 𝒫 𝐴 ∈ 𝒫 𝑥 ↔ ∀ 𝑦 ( ∀ 𝑧 ( 𝑧𝑦𝑧𝐴 ) → 𝑦𝑥 ) )
17 16 exbii ( ∃ 𝑥 𝒫 𝐴 ∈ 𝒫 𝑥 ↔ ∃ 𝑥𝑦 ( ∀ 𝑧 ( 𝑧𝑦𝑧𝐴 ) → 𝑦𝑥 ) )
18 8 17 bitri ( 𝒫 𝐴 ∈ V ↔ ∃ 𝑥𝑦 ( ∀ 𝑧 ( 𝑧𝑦𝑧𝐴 ) → 𝑦𝑥 ) )