Theorem flddivrng 38913

Description: Obsolete theorem, use flddrngd 20987 instead. A field is a division ring. (Contributed by Jeff Madsen, 10-Jun-2010.) (Revised by Mario Carneiro, 15-Dec-2013.) (Proof modification is discouraged.) (New usage is discouraged.)

Assertion

Ref	Expression
flddivrng	⊢ (𝐾 ∈ Fld → 𝐾 ∈ DivRingOps)

Proof of Theorem flddivrng

Step	Hyp	Ref	Expression
1		df-fld 38906	. . 3 ⊢ Fld = (DivRingOps ∩ Com2)
2		inss1 4182	. . 3 ⊢ (DivRingOps ∩ Com2) ⊆ DivRingOps
3	1, 2	eqsstri 3977	. 2 ⊢ Fld ⊆ DivRingOps
4	3	sseli 3927	1 ⊢ (𝐾 ∈ Fld → 𝐾 ∈ DivRingOps)

This proof depends on syntax axioms:   → wi 4   ∈ wcel 2145   ∩ cin 3898  DivRingOpscdrng 38862  Com2ccm2 38903  Fldcfld 38905

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 2733

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2740  df-cleq 2753  df-clel 2836  df-v 3453  df-in 3906  df-ss 3916  df-fld 38906

This theorem is used by:  isfld2  38919  isfldidl  38982