Theorem poss 5589

Description: Subset theorem for the partial ordering predicate. (Contributed by NM, 27-Mar-1997.) (Proof shortened by Mario Carneiro, 18-Nov-2016.)

Assertion

Ref	Expression
poss	⊢ (𝐴 ⊆ 𝐵 → (𝑅 Po 𝐵 → 𝑅 Po 𝐴))

Proof of Theorem poss

Dummy variables 𝑥 𝑦 𝑧 are mutually distinct and distinct from all other variables.

Step	Hyp	Ref	Expression
1		ssralv 4049	. . 3 ⊢ (𝐴 ⊆ 𝐵 → (∀𝑥 ∈ 𝐵 ∀𝑦 ∈ 𝐵 ∀𝑧 ∈ 𝐵 (¬ 𝑥𝑅𝑥 ∧ ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧)) → ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 ∀𝑧 ∈ 𝐵 (¬ 𝑥𝑅𝑥 ∧ ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧))))
2		ss2ralv 4051	. . . 4 ⊢ (𝐴 ⊆ 𝐵 → (∀𝑦 ∈ 𝐵 ∀𝑧 ∈ 𝐵 (¬ 𝑥𝑅𝑥 ∧ ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧)) → ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 (¬ 𝑥𝑅𝑥 ∧ ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧))))
3	2	ralimdv 3169	. . 3 ⊢ (𝐴 ⊆ 𝐵 → (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 ∀𝑧 ∈ 𝐵 (¬ 𝑥𝑅𝑥 ∧ ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧)) → ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 (¬ 𝑥𝑅𝑥 ∧ ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧))))
4	1, 3	syld 47	. 2 ⊢ (𝐴 ⊆ 𝐵 → (∀𝑥 ∈ 𝐵 ∀𝑦 ∈ 𝐵 ∀𝑧 ∈ 𝐵 (¬ 𝑥𝑅𝑥 ∧ ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧)) → ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 (¬ 𝑥𝑅𝑥 ∧ ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧))))
5		df-po 5587	. 2 ⊢ (𝑅 Po 𝐵 ↔ ∀𝑥 ∈ 𝐵 ∀𝑦 ∈ 𝐵 ∀𝑧 ∈ 𝐵 (¬ 𝑥𝑅𝑥 ∧ ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧)))
6		df-po 5587	. 2 ⊢ (𝑅 Po 𝐴 ↔ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 (¬ 𝑥𝑅𝑥 ∧ ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧)))
7	4, 5, 6	3imtr4g 295	1 ⊢ (𝐴 ⊆ 𝐵 → (𝑅 Po 𝐵 → 𝑅 Po 𝐴))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 396  ∀wral 3061   ⊆ wss 3947   class class class wbr 5147   Po wpo 5585

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-ext 2703

This theorem depends on definitions:  df-bi 206  df-an 397  df-tru 1544  df-ex 1782  df-sb 2068  df-clab 2710  df-cleq 2724  df-clel 2810  df-ral 3062  df-v 3476  df-in 3954  df-ss 3964  df-po 5587

This theorem is referenced by:  poeq2  5591  soss  5607  frpomin  6338  fprlem1  8281  swoso  8732  frfi  9284  wemapsolem  9541  fin23lem27  10319  zorn2lem6  10492  xrge0iifiso  32903  incsequz2  36605