Theorem rabsssn 4600

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 3147	. . 3 ⊢ {𝑥 ∈ 𝑉 ∣ 𝜑} = {𝑥 ∣ (𝑥 ∈ 𝑉 ∧ 𝜑)}
2		df-sn 4561	. . 3 ⊢ {𝑋} = {𝑥 ∣ 𝑥 = 𝑋}
3	1, 2	sseq12i 3996	. 2 ⊢ ({𝑥 ∈ 𝑉 ∣ 𝜑} ⊆ {𝑋} ↔ {𝑥 ∣ (𝑥 ∈ 𝑉 ∧ 𝜑)} ⊆ {𝑥 ∣ 𝑥 = 𝑋})
4		ss2ab 4038	. 2 ⊢ ({𝑥 ∣ (𝑥 ∈ 𝑉 ∧ 𝜑)} ⊆ {𝑥 ∣ 𝑥 = 𝑋} ↔ ∀𝑥((𝑥 ∈ 𝑉 ∧ 𝜑) → 𝑥 = 𝑋))
5		impexp 453	. . . 4 ⊢ (((𝑥 ∈ 𝑉 ∧ 𝜑) → 𝑥 = 𝑋) ↔ (𝑥 ∈ 𝑉 → (𝜑 → 𝑥 = 𝑋)))
6	5	albii 1816	. . 3 ⊢ (∀𝑥((𝑥 ∈ 𝑉 ∧ 𝜑) → 𝑥 = 𝑋) ↔ ∀𝑥(𝑥 ∈ 𝑉 → (𝜑 → 𝑥 = 𝑋)))
7		df-ral 3143	. . 3 ⊢ (∀𝑥 ∈ 𝑉 (𝜑 → 𝑥 = 𝑋) ↔ ∀𝑥(𝑥 ∈ 𝑉 → (𝜑 → 𝑥 = 𝑋)))
8	6, 7	bitr4i 280	. 2 ⊢ (∀𝑥((𝑥 ∈ 𝑉 ∧ 𝜑) → 𝑥 = 𝑋) ↔ ∀𝑥 ∈ 𝑉 (𝜑 → 𝑥 = 𝑋))
9	3, 4, 8	3bitri 299	1 ⊢ ({𝑥 ∈ 𝑉 ∣ 𝜑} ⊆ {𝑋} ↔ ∀𝑥 ∈ 𝑉 (𝜑 → 𝑥 = 𝑋))

Syntax hints:   → wi 4   ↔ wb 208   ∧ wa 398  ∀wal 1531   = wceq 1533   ∈ wcel 2110  {cab 2799  ∀wral 3138  {crab 3142   ⊆ wss 3935  {csn 4560

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1907  ax-6 1966  ax-7 2011  ax-8 2112  ax-9 2120  ax-10 2141  ax-11 2157  ax-12 2173  ax-ext 2793

This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-tru 1536  df-ex 1777  df-nf 1781  df-sb 2066  df-clab 2800  df-cleq 2814  df-clel 2893  df-nfc 2963  df-ral 3143  df-rab 3147  df-in 3942  df-ss 3951  df-sn 4561

This theorem is referenced by:  suppmptcfin  44421  linc1  44474