Theorem frss 2937

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 2082	. . . . . 6 |- (x (_ A -> (A (_ B -> x (_ B))
2	1	com12 11	. . . . 5 |- (A (_ B -> (x (_ A -> x (_ B))
3	2	anim1d 563	. . . 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 1293	. 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 2935	. 2 |- (R Fr B <-> A.x((x (_ B /\ x =/= (/)) -> E.y e. x (x i^i {z | zRy}) = (/)))
7		dffr2 2935	. 2 |- (R Fr A <-> A.x((x (_ A /\ x =/= (/)) -> E.y e. x (x i^i {z | zRy}) = (/)))
8	5, 6, 7	3imtr4g 556	1 |- (A (_ B -> (R Fr B -> R Fr A))

Syntax hints:   -> wi 3   /\ wa 223  A.wal 958   = wceq 960  {cab 1469   =/= wne 1592  E.wrex 1653   i^i cin 2057   (_ wss 2058  (/)c0 2291   class class class wbr 2634   Fr wfr 2931

This theorem is referenced by:  freq2 2939  wess 2952

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 966  ax-gen 967  ax-8 968  ax-10 970  ax-12 972  ax-17 975  ax-4 977  ax-5o 979  ax-6o 982  ax-9o 1127  ax-10o 1144  ax-16 1214  ax-11o 1222  ax-ext 1464

This theorem depends on definitions:  df-bi 147  df-or 224  df-an 225  df-ex 985  df-sb 1176  df-clab 1470  df-cleq 1475  df-clel 1478  df-ral 1656  df-rex 1657  df-v 1819  df-dif 2060  df-un 2061  df-in 2062  df-ss 2064  df-nul 2292  df-sn 2424  df-pr 2425  df-op 2428  df-br 2635  df-fr 2933

Copyright terms: Public domain