Theorem gruelss 10216

Description: A Grothendieck universe is transitive, so each element is a subset of the universe. (Contributed by Mario Carneiro, 9-Jun-2013.)

Assertion

Ref	Expression
gruelss	⊢ ((𝑈 ∈ Univ ∧ 𝐴 ∈ 𝑈) → 𝐴 ⊆ 𝑈)

Proof of Theorem gruelss

Step	Hyp	Ref	Expression
1		grutr 10215	. 2 ⊢ (𝑈 ∈ Univ → Tr 𝑈)
2		trss 5181	. . 3 ⊢ (Tr 𝑈 → (𝐴 ∈ 𝑈 → 𝐴 ⊆ 𝑈))
3	2	imp 409	. 2 ⊢ ((Tr 𝑈 ∧ 𝐴 ∈ 𝑈) → 𝐴 ⊆ 𝑈)
4	1, 3	sylan 582	1 ⊢ ((𝑈 ∈ Univ ∧ 𝐴 ∈ 𝑈) → 𝐴 ⊆ 𝑈)

Syntax hints:   → wi 4   ∧ wa 398   ∈ wcel 2114   ⊆ wss 3936  Tr wtr 5172  Univcgru 10212

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1970  ax-7 2015  ax-8 2116  ax-9 2124  ax-10 2145  ax-11 2161  ax-12 2177  ax-ext 2793

This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-3an 1085  df-tru 1540  df-ex 1781  df-nf 1785  df-sb 2070  df-clab 2800  df-cleq 2814  df-clel 2893  df-nfc 2963  df-ral 3143  df-rab 3147  df-v 3496  df-dif 3939  df-un 3941  df-in 3943  df-ss 3952  df-nul 4292  df-if 4468  df-sn 4568  df-pr 4570  df-op 4574  df-uni 4839  df-br 5067  df-tr 5173  df-iota 6314  df-fv 6363  df-ov 7159  df-gru 10213

This theorem is referenced by:  gruss  10218  gruuni  10222  gruel  10225  grur1a  10241  grur1  10242