Theorem ssrecnpr 45134

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 4611	. 2 ⊢ (𝑆 ∈ {ℝ, ℂ} → (𝑆 = ℝ ∨ 𝑆 = ℂ))
2		eqimss2 3993	. . 3 ⊢ (𝑆 = ℝ → ℝ ⊆ 𝑆)
3		ax-resscn 11184	. . . 4 ⊢ ℝ ⊆ ℂ
4		sseq2 3960	. . . 4 ⊢ (𝑆 = ℂ → (ℝ ⊆ 𝑆 ↔ ℝ ⊆ ℂ))
5	3, 4	mpbiri 261	. . 3 ⊢ (𝑆 = ℂ → ℝ ⊆ 𝑆)
6	2, 5	jaoi 871	. 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 861   = wceq 1570   ∈ wcel 2145   ⊆ wss 3902  {cpr 4589  ℂcc 11125  ℝcr 11126

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 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-ext 2734  ax-resscn 11184

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2741  df-cleq 2754  df-clel 2837  df-v 3455  df-un 3907  df-ss 3919  df-sn 4588  df-pr 4590

This theorem is used by: (None)