Theorem ssrecnpr 40130

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 4488	. 2 ⊢ (𝑆 ∈ {ℝ, ℂ} → (𝑆 = ℝ ∨ 𝑆 = ℂ))
2		eqimss2 3940	. . 3 ⊢ (𝑆 = ℝ → ℝ ⊆ 𝑆)
3		ax-resscn 10429	. . . 4 ⊢ ℝ ⊆ ℂ
4		sseq2 3909	. . . 4 ⊢ (𝑆 = ℂ → (ℝ ⊆ 𝑆 ↔ ℝ ⊆ ℂ))
5	3, 4	mpbiri 259	. . 3 ⊢ (𝑆 = ℂ → ℝ ⊆ 𝑆)
6	2, 5	jaoi 852	. 2 ⊢ ((𝑆 = ℝ ∨ 𝑆 = ℂ) → ℝ ⊆ 𝑆)
7	1, 6	syl 17	1 ⊢ (𝑆 ∈ {ℝ, ℂ} → ℝ ⊆ 𝑆)

Syntax hints:   → wi 4   ∨ wo 842   = wceq 1520   ∈ wcel 2079   ⊆ wss 3854  {cpr 4468  ℂcc 10370  ℝcr 10371

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1775  ax-4 1789  ax-5 1886  ax-6 1945  ax-7 1990  ax-8 2081  ax-9 2089  ax-10 2110  ax-11 2124  ax-12 2139  ax-ext 2767  ax-resscn 10429

This theorem depends on definitions:  df-bi 208  df-an 397  df-or 843  df-tru 1523  df-ex 1760  df-nf 1764  df-sb 2041  df-clab 2774  df-cleq 2786  df-clel 2861  df-nfc 2933  df-v 3434  df-un 3859  df-in 3861  df-ss 3869  df-sn 4467  df-pr 4469

This theorem is referenced by: (None)