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 = (/)) -> E.y e. x (x i^i {z | zRy}) = (/)) -> ((x (_ A /\ -. x = (/)) -> E.y e. x (x i^i {z | zRy}) = (/))))
5	4	19.20dv 946	. 2 |- (A (_ B -> (A.x((x (_ B /\ -. x = (/)) -> E.y e. x (x i^i {z | zRy}) = (/)) -> A.x((x (_ A /\ -. x = (/)) -> E.y e. x (x i^i {z | zRy}) = (/))))
6		dffr2 2171	. 2 |- (R Fr B <-> A.x((x (_ B /\ -. x = (/)) -> E.y e. x (x i^i {z | zRy}) = (/)))
7		dffr2 2171	. 2 |- (R Fr A <-> A.x((x (_ A /\ -. x = (/)) -> E.y e. x (x i^i {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  A.wal 672  {cab 1090   = wceq 1091  E.wrex 1202   i^i 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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