Theorem axrep3 5242

Description: Axiom of Replacement slightly strengthened from axrep2 5241; 𝑤 may occur free in 𝜑. (Contributed by NM, 2-Jan-1997.) Remove dependency on ax-13 2404. (Revised by BJ, 31-May-2019.)

Assertion

Ref	Expression
axrep3	⊢ ∃𝑥(∃𝑦∀𝑧(𝜑 → 𝑧 = 𝑦) → ∀𝑧(𝑧 ∈ 𝑥 ↔ ∃𝑥(𝑥 ∈ 𝑤 ∧ ∀𝑦𝜑)))

Distinct variable group:   𝑥,𝑤,𝑦,𝑧

Allowed substitution hints:   𝜑(𝑥, 𝑦, 𝑧, 𝑤)

Proof of Theorem axrep3

Step	Hyp	Ref	Expression
1		nfe1 2190	. . . 4 ⊢ Ⅎ𝑦∃𝑦∀𝑧(𝜑 → 𝑧 = 𝑦)
2		nfv 1949	. . . . . 6 ⊢ Ⅎ𝑦 𝑧 ∈ 𝑥
3		nfv 1949	. . . . . . . 8 ⊢ Ⅎ𝑦 𝑥 ∈ 𝑤
4		nfa1 2191	. . . . . . . 8 ⊢ Ⅎ𝑦∀𝑦𝜑
5	3, 4	nfan 1934	. . . . . . 7 ⊢ Ⅎ𝑦(𝑥 ∈ 𝑤 ∧ ∀𝑦𝜑)
6	5	nfex 2358	. . . . . 6 ⊢ Ⅎ𝑦∃𝑥(𝑥 ∈ 𝑤 ∧ ∀𝑦𝜑)
7	2, 6	nfbi 1938	. . . . 5 ⊢ Ⅎ𝑦(𝑧 ∈ 𝑥 ↔ ∃𝑥(𝑥 ∈ 𝑤 ∧ ∀𝑦𝜑))
8	7	nfal 2357	. . . 4 ⊢ Ⅎ𝑦∀𝑧(𝑧 ∈ 𝑥 ↔ ∃𝑥(𝑥 ∈ 𝑤 ∧ ∀𝑦𝜑))
9	1, 8	nfim 1931	. . 3 ⊢ Ⅎ𝑦(∃𝑦∀𝑧(𝜑 → 𝑧 = 𝑦) → ∀𝑧(𝑧 ∈ 𝑥 ↔ ∃𝑥(𝑥 ∈ 𝑤 ∧ ∀𝑦𝜑)))
10	9	nfex 2358	. 2 ⊢ Ⅎ𝑦∃𝑥(∃𝑦∀𝑧(𝜑 → 𝑧 = 𝑦) → ∀𝑧(𝑧 ∈ 𝑥 ↔ ∃𝑥(𝑥 ∈ 𝑤 ∧ ∀𝑦𝜑)))
11		axreplem 5240	. 2 ⊢ (𝑦 = 𝑤 → (∃𝑥(∃𝑦∀𝑧(𝜑 → 𝑧 = 𝑦) → ∀𝑧(𝑧 ∈ 𝑥 ↔ ∃𝑥(𝑥 ∈ 𝑦 ∧ ∀𝑦𝜑))) ↔ ∃𝑥(∃𝑦∀𝑧(𝜑 → 𝑧 = 𝑦) → ∀𝑧(𝑧 ∈ 𝑥 ↔ ∃𝑥(𝑥 ∈ 𝑤 ∧ ∀𝑦𝜑)))))
12		axrep2 5241	. 2 ⊢ ∃𝑥(∃𝑦∀𝑧(𝜑 → 𝑧 = 𝑦) → ∀𝑧(𝑧 ∈ 𝑥 ↔ ∃𝑥(𝑥 ∈ 𝑦 ∧ ∀𝑦𝜑)))
13	10, 11, 12	chvarfv 2280	1 ⊢ ∃𝑥(∃𝑦∀𝑧(𝜑 → 𝑧 = 𝑦) → ∀𝑧(𝑧 ∈ 𝑥 ↔ ∃𝑥(𝑥 ∈ 𝑤 ∧ ∀𝑦𝜑)))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401  ∀wal 1568  ∃wex 1812

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 1945  ax-6 2002  ax-7 2043  ax-9 2157  ax-10 2180  ax-11 2197  ax-12 2216  ax-rep 5238

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-ex 1813  df-nf 1817

This theorem is used by: (None)