Theorem dfssf 3899

Description: Equivalence for subclass relation, using bound-variable hypotheses instead of distinct variable conditions. (Contributed by NM, 3-Jul-1994.) (Revised by Andrew Salmon, 27-Aug-2011.) Avoid ax-13 2401. (Revised by GG, 19-May-2023.)

Hypotheses

Ref	Expression
dfssf.1	⊢ Ⅎ𝑥𝐴
dfssf.2	⊢ Ⅎ𝑥𝐵

Assertion

Ref	Expression
dfssf	⊢ (𝐴 ⊆ 𝐵 ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))

Proof of Theorem dfssf

Dummy variable 𝑧 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		df-ss 3896	. 2 ⊢ (𝐴 ⊆ 𝐵 ↔ ∀𝑧(𝑧 ∈ 𝐴 → 𝑧 ∈ 𝐵))
2		dfssf.1	. . . . 5 ⊢ Ⅎ𝑥𝐴
3	2	nfcri 2914	. . . 4 ⊢ Ⅎ𝑥 𝑧 ∈ 𝐴
4		dfssf.2	. . . . 5 ⊢ Ⅎ𝑥𝐵
5	4	nfcri 2914	. . . 4 ⊢ Ⅎ𝑥 𝑧 ∈ 𝐵
6	3, 5	nfim 1928	. . 3 ⊢ Ⅎ𝑥(𝑧 ∈ 𝐴 → 𝑧 ∈ 𝐵)
7		nfv 1946	. . 3 ⊢ Ⅎ𝑧(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵)
8		eleq1w 2843	. . . 4 ⊢ (𝑧 = 𝑥 → (𝑧 ∈ 𝐴 ↔ 𝑥 ∈ 𝐴))
9		eleq1w 2843	. . . 4 ⊢ (𝑧 = 𝑥 → (𝑧 ∈ 𝐵 ↔ 𝑥 ∈ 𝐵))
10	8, 9	imbi12d 347	. . 3 ⊢ (𝑧 = 𝑥 → ((𝑧 ∈ 𝐴 → 𝑧 ∈ 𝐵) ↔ (𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵)))
11	6, 7, 10	cbvalv1 2370	. 2 ⊢ (∀𝑧(𝑧 ∈ 𝐴 → 𝑧 ∈ 𝐵) ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
12	1, 11	bitri 278	1 ⊢ (𝐴 ⊆ 𝐵 ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209  ∀wal 1567   ∈ wcel 2144  Ⅎwnfc 2907   ⊆ wss 3895

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-11 2193  ax-12 2212

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1812  df-nf 1816  df-clel 2835  df-nfc 2909  df-ss 3896

This theorem is used by:  dfss3f  3900  ssrd  3913  ssrmof  3976  ss2ab  3986  rankval4b  9884  rankval4  9888  rabexgfGS  33143  ballotth  35217  dvcosre  47016  itgsinexplem1  47058