Theorem ssrecnpr 45046

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 4612	. 2 ⊢ (𝑆 ∈ {ℝ, ℂ} → (𝑆 = ℝ ∨ 𝑆 = ℂ))
2		eqimss2 3995	. . 3 ⊢ (𝑆 = ℝ → ℝ ⊆ 𝑆)
3		ax-resscn 11163	. . . 4 ⊢ ℝ ⊆ ℂ
4		sseq2 3962	. . . 4 ⊢ (𝑆 = ℂ → (ℝ ⊆ 𝑆 ↔ ℝ ⊆ ℂ))
5	3, 4	mpbiri 261	. . 3 ⊢ (𝑆 = ℂ → ℝ ⊆ 𝑆)
6	2, 5	jaoi 870	. 2 ⊢ ((𝑆 = ℝ ∨ 𝑆 = ℂ) → ℝ ⊆ 𝑆)
7	1, 6	syl 18	1 ⊢ (𝑆 ∈ {ℝ, ℂ} → ℝ ⊆ 𝑆)

This proof depends on syntax axioms:   → wi 4   ∨ wo 860   = wceq 1569   ∈ wcel 2142   ⊆ wss 3904  {cpr 4590  ℂcc 11104  ℝcr 11105

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1824  ax-4 1838  ax-5 1939  ax-6 1996  ax-7 2037  ax-8 2144  ax-9 2152  ax-ext 2734  ax-resscn 11163

This proof depends on definitions:  df-bi 210  df-an 401  df-or 861  df-tru 1572  df-ex 1809  df-sb 2096  df-clab 2741  df-cleq 2754  df-clel 2837  df-v 3456  df-un 3909  df-ss 3921  df-sn 4589  df-pr 4591

This theorem is used by: (None)