Theorem ssrecnpr 45368

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 4607	. 2 ⊢ (𝑆 ∈ {ℝ, ℂ} → (𝑆 = ℝ ∨ 𝑆 = ℂ))
2		eqimss2 3967	. . 3 ⊢ (𝑆 = ℝ → ℝ ⊆ 𝑆)
3		ax-resscn 11258	. . . 4 ⊢ ℝ ⊆ ℂ
4		sseq2 3934	. . . 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 1569   ∈ wcel 2144   ⊆ wss 3895  {cpr 4585  ℂcc 11199  ℝcr 11200

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-ext 2732  ax-resscn 11258

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1572  df-ex 1812  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-v 3452  df-ss 3896  df-un 4151  df-sn 4584  df-pr 4586

This theorem is used by: (None)