Theorem poss 5568

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 3982	. . 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 3984	. . . 4 ⊢ (𝐴 ⊆ 𝐵 → (∀𝑦 ∈ 𝐵 ∀𝑧 ∈ 𝐵 (¬ 𝑥𝑅𝑥 ∧ ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧)) → ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 (¬ 𝑥𝑅𝑥 ∧ ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧))))
3	2	ralimdv 3180	. . 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 48	. 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 5566	. 2 ⊢ (𝑅 Po 𝐵 ↔ ∀𝑥 ∈ 𝐵 ∀𝑦 ∈ 𝐵 ∀𝑧 ∈ 𝐵 (¬ 𝑥𝑅𝑥 ∧ ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧)))
6		df-po 5566	. 2 ⊢ (𝑅 Po 𝐴 ↔ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ∀𝑧 ∈ 𝐴 (¬ 𝑥𝑅𝑥 ∧ ((𝑥𝑅𝑦 ∧ 𝑦𝑅𝑧) → 𝑥𝑅𝑧)))
7	4, 5, 6	3imtr4g 299	1 ⊢ (𝐴 ⊆ 𝐵 → (𝑅 Po 𝐵 → 𝑅 Po 𝐴))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ∧ wa 401  ∀wral 3080   ⊆ wss 3900   class class class wbr 5108   Po wpo 5564

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 1945

This proof depends on definitions:  df-bi 210  df-an 402  df-ral 3081  df-ss 3901  df-po 5566

This theorem is used by:  poeq2  5570  soss  5586  frpomin  6369  fprlem1  8356  swoso  8790  frfi  9314  wemapsolem  9581  fin23lem27  10445  zorn2lem6  10618  chnso  18848  xrge0iifiso  34678  incsequz2  38949  chnsuslle  48164