Theorem ssrecnpr 41815

Description: ℝ is a subset of both ℝ and ℂ. (Contributed by Steve Rodriguez, 22-Nov-2015.)

Assertion

Ref	Expression
ssrecnpr	⊢ (𝑆 ∈ {ℝ, ℂ} → ℝ ⊆ 𝑆)

Proof of Theorem ssrecnpr

Step	Hyp	Ref	Expression
1		elpri 4580	. 2 ⊢ (𝑆 ∈ {ℝ, ℂ} → (𝑆 = ℝ ∨ 𝑆 = ℂ))
2		eqimss2 3974	. . 3 ⊢ (𝑆 = ℝ → ℝ ⊆ 𝑆)
3		ax-resscn 10859	. . . 4 ⊢ ℝ ⊆ ℂ
4		sseq2 3943	. . . 4 ⊢ (𝑆 = ℂ → (ℝ ⊆ 𝑆 ↔ ℝ ⊆ ℂ))
5	3, 4	mpbiri 257	. . 3 ⊢ (𝑆 = ℂ → ℝ ⊆ 𝑆)
6	2, 5	jaoi 853	. 2 ⊢ ((𝑆 = ℝ ∨ 𝑆 = ℂ) → ℝ ⊆ 𝑆)
7	1, 6	syl 17	1 ⊢ (𝑆 ∈ {ℝ, ℂ} → ℝ ⊆ 𝑆)

Syntax hints:   → wi 4   ∨ wo 843   = wceq 1539   ∈ wcel 2108   ⊆ wss 3883  {cpr 4560  ℂcc 10800  ℝcr 10801

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1799  ax-4 1813  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2110  ax-9 2118  ax-ext 2709  ax-resscn 10859

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 844  df-tru 1542  df-ex 1784  df-sb 2069  df-clab 2716  df-cleq 2730  df-clel 2817  df-v 3424  df-un 3888  df-in 3890  df-ss 3900  df-sn 4559  df-pr 4561

This theorem is referenced by: (None)