Theorem frss

Description: Subset theorem for the well-founded predicate. Exercise 1 of TakeutiZaring p. 31. (Contributed by NM, 3-Apr-1994) (Proof shortened by Andrew Salmon, 25-Jul-2011)

	Ref	Expression
Assertion	frss	⊢ ( 𝐴 ⊆ 𝐵 → ( 𝑅 Fr 𝐵 → 𝑅 Fr 𝐴 ) )

Proof

Step	Hyp	Ref	Expression
1		sstr2	⊢ ( 𝑥 ⊆ 𝐴 → ( 𝐴 ⊆ 𝐵 → 𝑥 ⊆ 𝐵 ) )
2	1	com12	⊢ ( 𝐴 ⊆ 𝐵 → ( 𝑥 ⊆ 𝐴 → 𝑥 ⊆ 𝐵 ) )
3	2	anim1d	⊢ ( 𝐴 ⊆ 𝐵 → ( ( 𝑥 ⊆ 𝐴 ∧ 𝑥 ≠ ∅ ) → ( 𝑥 ⊆ 𝐵 ∧ 𝑥 ≠ ∅ ) ) )
4	3	imim1d	⊢ ( 𝐴 ⊆ 𝐵 → ( ( ( 𝑥 ⊆ 𝐵 ∧ 𝑥 ≠ ∅ ) → ∃ 𝑦 ∈ 𝑥 ∀ 𝑧 ∈ 𝑥 ¬ 𝑧 𝑅 𝑦 ) → ( ( 𝑥 ⊆ 𝐴 ∧ 𝑥 ≠ ∅ ) → ∃ 𝑦 ∈ 𝑥 ∀ 𝑧 ∈ 𝑥 ¬ 𝑧 𝑅 𝑦 ) ) )
5	4	alimdv	⊢ ( 𝐴 ⊆ 𝐵 → ( ∀ 𝑥 ( ( 𝑥 ⊆ 𝐵 ∧ 𝑥 ≠ ∅ ) → ∃ 𝑦 ∈ 𝑥 ∀ 𝑧 ∈ 𝑥 ¬ 𝑧 𝑅 𝑦 ) → ∀ 𝑥 ( ( 𝑥 ⊆ 𝐴 ∧ 𝑥 ≠ ∅ ) → ∃ 𝑦 ∈ 𝑥 ∀ 𝑧 ∈ 𝑥 ¬ 𝑧 𝑅 𝑦 ) ) )
6		df-fr	⊢ ( 𝑅 Fr 𝐵 ↔ ∀ 𝑥 ( ( 𝑥 ⊆ 𝐵 ∧ 𝑥 ≠ ∅ ) → ∃ 𝑦 ∈ 𝑥 ∀ 𝑧 ∈ 𝑥 ¬ 𝑧 𝑅 𝑦 ) )
7		df-fr	⊢ ( 𝑅 Fr 𝐴 ↔ ∀ 𝑥 ( ( 𝑥 ⊆ 𝐴 ∧ 𝑥 ≠ ∅ ) → ∃ 𝑦 ∈ 𝑥 ∀ 𝑧 ∈ 𝑥 ¬ 𝑧 𝑅 𝑦 ) )
8	5 6 7	3imtr4g	⊢ ( 𝐴 ⊆ 𝐵 → ( 𝑅 Fr 𝐵 → 𝑅 Fr 𝐴 ) )