Theorem frss 5649

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.)

Assertion

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

Proof of Theorem frss

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

Step	Hyp	Ref	Expression
1		sstr2 3990	. . . . . 6 ⊢ (𝑥 ⊆ 𝐴 → (𝐴 ⊆ 𝐵 → 𝑥 ⊆ 𝐵))
2	1	com12 32	. . . . 5 ⊢ (𝐴 ⊆ 𝐵 → (𝑥 ⊆ 𝐴 → 𝑥 ⊆ 𝐵))
3	2	anim1d 611	. . . 4 ⊢ (𝐴 ⊆ 𝐵 → ((𝑥 ⊆ 𝐴 ∧ 𝑥 ≠ ∅) → (𝑥 ⊆ 𝐵 ∧ 𝑥 ≠ ∅)))
4	3	imim1d 82	. . 3 ⊢ (𝐴 ⊆ 𝐵 → (((𝑥 ⊆ 𝐵 ∧ 𝑥 ≠ ∅) → ∃𝑦 ∈ 𝑥 ∀𝑧 ∈ 𝑥 ¬ 𝑧𝑅𝑦) → ((𝑥 ⊆ 𝐴 ∧ 𝑥 ≠ ∅) → ∃𝑦 ∈ 𝑥 ∀𝑧 ∈ 𝑥 ¬ 𝑧𝑅𝑦)))
5	4	alimdv 1916	. 2 ⊢ (𝐴 ⊆ 𝐵 → (∀𝑥((𝑥 ⊆ 𝐵 ∧ 𝑥 ≠ ∅) → ∃𝑦 ∈ 𝑥 ∀𝑧 ∈ 𝑥 ¬ 𝑧𝑅𝑦) → ∀𝑥((𝑥 ⊆ 𝐴 ∧ 𝑥 ≠ ∅) → ∃𝑦 ∈ 𝑥 ∀𝑧 ∈ 𝑥 ¬ 𝑧𝑅𝑦)))
6		df-fr 5637	. 2 ⊢ (𝑅 Fr 𝐵 ↔ ∀𝑥((𝑥 ⊆ 𝐵 ∧ 𝑥 ≠ ∅) → ∃𝑦 ∈ 𝑥 ∀𝑧 ∈ 𝑥 ¬ 𝑧𝑅𝑦))
7		df-fr 5637	. 2 ⊢ (𝑅 Fr 𝐴 ↔ ∀𝑥((𝑥 ⊆ 𝐴 ∧ 𝑥 ≠ ∅) → ∃𝑦 ∈ 𝑥 ∀𝑧 ∈ 𝑥 ¬ 𝑧𝑅𝑦))
8	5, 6, 7	3imtr4g 296	1 ⊢ (𝐴 ⊆ 𝐵 → (𝑅 Fr 𝐵 → 𝑅 Fr 𝐴))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 395  ∀wal 1538   ≠ wne 2940  ∀wral 3061  ∃wrex 3070   ⊆ wss 3951  ∅c0 4333   class class class wbr 5143   Fr wfr 5634

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910

This theorem depends on definitions:  df-bi 207  df-an 396  df-ss 3968  df-fr 5637

This theorem is referenced by:  freq2  5653  wess  5671  fprlem1  8325  frmin  9789  frrlem15  9797  tcfr  44980