Theorem frss 2173

Description: Subset theorem for the founded predicate. Exercise 1 of [TakeutiZaring] p. 31.

Assertion

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

Proof of Theorem frss

Step	Hyp	Ref	Expression
1		sstr2 1510	. . . . . 6 ⊢ (x ⊆ A → (A ⊆ B → x ⊆ B))
2	1	com12 13	. . . . 5 ⊢ (A ⊆ B → (x ⊆ A → x ⊆ B))
3	2	anim1d 432	. . . 4 ⊢ (A ⊆ B → ((x ⊆ A ∧ ¬ x = ∅) → (x ⊆ B ∧ ¬ x = ∅)))
4	3	syl4d 28	. . 3 ⊢ (A ⊆ B → (((x ⊆ B ∧ ¬ x = ∅) → ∃y ∈ x (x ∩ {z∣zRy}) = ∅) → ((x ⊆ A ∧ ¬ x = ∅) → ∃y ∈ x (x ∩ {z∣zRy}) = ∅)))
5	4	19.20dv 946	. 2 ⊢ (A ⊆ B → (∀x((x ⊆ B ∧ ¬ x = ∅) → ∃y ∈ x (x ∩ {z∣zRy}) = ∅) → ∀x((x ⊆ A ∧ ¬ x = ∅) → ∃y ∈ x (x ∩ {z∣zRy}) = ∅)))
6		dffr2 2171	. 2 ⊢ (R Fr B ↔ ∀x((x ⊆ B ∧ ¬ x = ∅) → ∃y ∈ x (x ∩ {z∣zRy}) = ∅))
7		dffr2 2171	. 2 ⊢ (R Fr A ↔ ∀x((x ⊆ A ∧ ¬ x = ∅) → ∃y ∈ x (x ∩ {z∣zRy}) = ∅))
8	5, 6, 7	3imtr4g 426	1 ⊢ (A ⊆ B → (R Fr B → R Fr A))

Syntax hints:  ¬ wn 1   → wi 2   ∧ wa 196  ∀wal 672  {cab 1090   = wceq 1091  ∃wrex 1202   ∩ cin 1486   ⊆ wss 1487  ∅c0 1707   class class class wbr 2054   Fr wfr 2061

This theorem is referenced by:  freq2 2175  wess 2188

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802  ax-16 922  ax-17 925  ax-ext 1074

This theorem depends on definitions:  df-bi 128  df-or 197  df-an 198  df-ex 679  df-sb 853  df-clab 1093  df-cleq 1097  df-clel 1099  df-ral 1205  df-rex 1206  df-v 1349  df-dif 1489  df-un 1490  df-in 1491  df-ss 1492  df-nul 1708  df-sn 1811  df-pr 1812  df-op 1815  df-br 2063  df-fr 2169

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