Theorem gruelss 10215

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 10214	. 2 ⊢ (𝑈 ∈ Univ → Tr 𝑈)
2		trss 5180	. . 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 2110   ⊆ wss 3935  Tr wtr 5171  Univcgru 10211

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1907  ax-6 1966  ax-7 2011  ax-8 2112  ax-9 2120  ax-10 2141  ax-11 2157  ax-12 2173  ax-ext 2793

This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-3an 1085  df-tru 1536  df-ex 1777  df-nf 1781  df-sb 2066  df-clab 2800  df-cleq 2814  df-clel 2893  df-nfc 2963  df-ral 3143  df-rab 3147  df-v 3496  df-dif 3938  df-un 3940  df-in 3942  df-ss 3951  df-nul 4291  df-if 4467  df-sn 4567  df-pr 4569  df-op 4573  df-uni 4838  df-br 5066  df-tr 5172  df-iota 6313  df-fv 6362  df-ov 7158  df-gru 10212

This theorem is referenced by:  gruss  10217  gruuni  10221  gruel  10224  grur1a  10240  grur1  10241