Theorem frss 2927

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 2074	. . . . . 6 |- (x (_ A -> (A (_ B -> x (_ B))
2	1	com12 11	. . . . 5 |- (A (_ B -> (x (_ A -> x (_ B))
3	2	anim1d 562	. . . 4 |- (A (_ B -> ((x (_ A /\ x =/= (/)) -> (x (_ B /\ x =/= (/))))
4	3	imim1d 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 1291	. 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 2925	. 2 |- (R Fr B <-> A.x((x (_ B /\ x =/= (/)) -> E.y e. x (x i^i {z | zRy}) = (/)))
7		dffr2 2925	. 2 |- (R Fr A <-> A.x((x (_ A /\ x =/= (/)) -> E.y e. x (x i^i {z | zRy}) = (/)))
8	5, 6, 7	3imtr4g 555	1 |- (A (_ B -> (R Fr B -> R Fr A))

Syntax hints:   -> wi 3   /\ wa 223  A.wal 956   = wceq 958  {cab 1466   =/= wne 1588  E.wrex 1649   i^i cin 2049   (_ wss 2050  (/)c0 2283   class class class wbr 2624   Fr wfr 2921

This theorem is referenced by:  freq2 2929  wess 2942

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 964  ax-gen 965  ax-8 966  ax-10 968  ax-12 970  ax-17 973  ax-4 975  ax-5o 977  ax-6o 980  ax-9o 1125  ax-10o 1142  ax-16 1212  ax-11o 1220  ax-ext 1462

This theorem depends on definitions:  df-bi 147  df-or 224  df-an 225  df-ex 983  df-sb 1174  df-clab 1467  df-cleq 1472  df-clel 1475  df-ral 1652  df-rex 1653  df-v 1815  df-dif 2052  df-un 2053  df-in 2054  df-ss 2056  df-nul 2284  df-sn 2416  df-pr 2417  df-op 2420  df-br 2625  df-fr 2923

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