Theorem poss

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

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

Proof

Step	Hyp	Ref	Expression
1		ssralv	⊢ ( 𝐴 ⊆ 𝐵 → ( ∀ 𝑥 ∈ 𝐵 ∀ 𝑦 ∈ 𝐵 ∀ 𝑧 ∈ 𝐵 ( ¬ 𝑥 𝑅 𝑥 ∧ ( ( 𝑥 𝑅 𝑦 ∧ 𝑦 𝑅 𝑧 ) → 𝑥 𝑅 𝑧 ) ) → ∀ 𝑥 ∈ 𝐴 ∀ 𝑦 ∈ 𝐵 ∀ 𝑧 ∈ 𝐵 ( ¬ 𝑥 𝑅 𝑥 ∧ ( ( 𝑥 𝑅 𝑦 ∧ 𝑦 𝑅 𝑧 ) → 𝑥 𝑅 𝑧 ) ) ) )
2		ss2ralv	⊢ ( 𝐴 ⊆ 𝐵 → ( ∀ 𝑦 ∈ 𝐵 ∀ 𝑧 ∈ 𝐵 ( ¬ 𝑥 𝑅 𝑥 ∧ ( ( 𝑥 𝑅 𝑦 ∧ 𝑦 𝑅 𝑧 ) → 𝑥 𝑅 𝑧 ) ) → ∀ 𝑦 ∈ 𝐴 ∀ 𝑧 ∈ 𝐴 ( ¬ 𝑥 𝑅 𝑥 ∧ ( ( 𝑥 𝑅 𝑦 ∧ 𝑦 𝑅 𝑧 ) → 𝑥 𝑅 𝑧 ) ) ) )
3	2	ralimdv	⊢ ( 𝐴 ⊆ 𝐵 → ( ∀ 𝑥 ∈ 𝐴 ∀ 𝑦 ∈ 𝐵 ∀ 𝑧 ∈ 𝐵 ( ¬ 𝑥 𝑅 𝑥 ∧ ( ( 𝑥 𝑅 𝑦 ∧ 𝑦 𝑅 𝑧 ) → 𝑥 𝑅 𝑧 ) ) → ∀ 𝑥 ∈ 𝐴 ∀ 𝑦 ∈ 𝐴 ∀ 𝑧 ∈ 𝐴 ( ¬ 𝑥 𝑅 𝑥 ∧ ( ( 𝑥 𝑅 𝑦 ∧ 𝑦 𝑅 𝑧 ) → 𝑥 𝑅 𝑧 ) ) ) )
4	1 3	syld	⊢ ( 𝐴 ⊆ 𝐵 → ( ∀ 𝑥 ∈ 𝐵 ∀ 𝑦 ∈ 𝐵 ∀ 𝑧 ∈ 𝐵 ( ¬ 𝑥 𝑅 𝑥 ∧ ( ( 𝑥 𝑅 𝑦 ∧ 𝑦 𝑅 𝑧 ) → 𝑥 𝑅 𝑧 ) ) → ∀ 𝑥 ∈ 𝐴 ∀ 𝑦 ∈ 𝐴 ∀ 𝑧 ∈ 𝐴 ( ¬ 𝑥 𝑅 𝑥 ∧ ( ( 𝑥 𝑅 𝑦 ∧ 𝑦 𝑅 𝑧 ) → 𝑥 𝑅 𝑧 ) ) ) )
5		df-po	⊢ ( 𝑅 Po 𝐵 ↔ ∀ 𝑥 ∈ 𝐵 ∀ 𝑦 ∈ 𝐵 ∀ 𝑧 ∈ 𝐵 ( ¬ 𝑥 𝑅 𝑥 ∧ ( ( 𝑥 𝑅 𝑦 ∧ 𝑦 𝑅 𝑧 ) → 𝑥 𝑅 𝑧 ) ) )
6		df-po	⊢ ( 𝑅 Po 𝐴 ↔ ∀ 𝑥 ∈ 𝐴 ∀ 𝑦 ∈ 𝐴 ∀ 𝑧 ∈ 𝐴 ( ¬ 𝑥 𝑅 𝑥 ∧ ( ( 𝑥 𝑅 𝑦 ∧ 𝑦 𝑅 𝑧 ) → 𝑥 𝑅 𝑧 ) ) )
7	4 5 6	3imtr4g	⊢ ( 𝐴 ⊆ 𝐵 → ( 𝑅 Po 𝐵 → 𝑅 Po 𝐴 ) )