Theorem rabsssn 4629

Description: Conditions for a restricted class abstraction to be a subset of a singleton, i.e. to be a singleton or the empty set. (Contributed by AV, 18-Apr-2019.)

Assertion

Ref	Expression
rabsssn	⊢ ({𝑥 ∈ 𝑉 ∣ 𝜑} ⊆ {𝑋} ↔ ∀𝑥 ∈ 𝑉 (𝜑 → 𝑥 = 𝑋))

Distinct variable group:   𝑥,𝑋

Allowed substitution hints:   𝜑(𝑥)   𝑉(𝑥)

Proof of Theorem rabsssn

Step	Hyp	Ref	Expression
1		df-rab 3417	. . 3 ⊢ {𝑥 ∈ 𝑉 ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ 𝑉 ∧ 𝜑)}
2		df-sn 4589	. . 3 ⊢ {𝑋} = {𝑥 ∣ 𝑥 = 𝑋}
3	1, 2	sseq12i 3943	. 2 ⊢ ({𝑥 ∈ 𝑉 ∣ 𝜑} ⊆ {𝑋} ↔ {𝑥 ∣ (𝑥 ∈ 𝑉 ∧ 𝜑)} ⊆ {𝑥 ∣ 𝑥 = 𝑋})
4		ss2ab 3989	. 2 ⊢ ({𝑥 ∣ (𝑥 ∈ 𝑉 ∧ 𝜑)} ⊆ {𝑥 ∣ 𝑥 = 𝑋} ↔ ∀𝑥((𝑥 ∈ 𝑉 ∧ 𝜑) → 𝑥 = 𝑋))
5		impexp 456	. . . 4 ⊢ (((𝑥 ∈ 𝑉 ∧ 𝜑) → 𝑥 = 𝑋) ↔ (𝑥 ∈ 𝑉 → (𝜑 → 𝑥 = 𝑋)))
6	5	albii 1852	. . 3 ⊢ (∀𝑥((𝑥 ∈ 𝑉 ∧ 𝜑) → 𝑥 = 𝑋) ↔ ∀𝑥(𝑥 ∈ 𝑉 → (𝜑 → 𝑥 = 𝑋)))
7		df-ral 3081	. . 3 ⊢ (∀𝑥 ∈ 𝑉 (𝜑 → 𝑥 = 𝑋) ↔ ∀𝑥(𝑥 ∈ 𝑉 → (𝜑 → 𝑥 = 𝑋)))
8	6, 7	bitr4i 281	. 2 ⊢ (∀𝑥((𝑥 ∈ 𝑉 ∧ 𝜑) → 𝑥 = 𝑋) ↔ ∀𝑥 ∈ 𝑉 (𝜑 → 𝑥 = 𝑋))
9	3, 4, 8	3bitri 300	1 ⊢ ({𝑥 ∈ 𝑉 ∣ 𝜑} ⊆ {𝑋} ↔ ∀𝑥 ∈ 𝑉 (𝜑 → 𝑥 = 𝑋))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401  ∀wal 1568   = wceq 1570   ∈ wcel 2147  {cab 2741  ∀wral 3080  {crab 3416   ⊆ wss 3900  {csn 4588

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-8 2149  ax-9 2157  ax-10 2180  ax-11 2197  ax-12 2216  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-ex 1813  df-nf 1817  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-nfc 2913  df-ral 3081  df-rab 3417  df-ss 3901  df-sn 4589

This theorem is used by:  constrfin  34445  0elscott  35836  suppmptcfin  49691  linc1  49740