Metamath Proof Explorer


Theorem dffrege76

Description: If from the two propositions that every result of an application of the procedure R to B has property f and that property f is hereditary in the R -sequence, it can be inferred, whatever f may be, that E has property f , then we say E follows B in the R -sequence. Definition 76 of Frege1879 p. 60.

Each of B , E and R must be sets. (Contributed by RP, 2-Jul-2020)

Ref Expression
Hypotheses frege76.b BU
frege76.e EV
frege76.r RW
Assertion dffrege76 fRhereditaryfaBRaafEfBt+RE

Proof

Step Hyp Ref Expression
1 frege76.b BU
2 frege76.e EV
3 frege76.r RW
4 brtrclfv2 BUEVRWBt+REEf|RBff
5 1 2 3 4 mp3an Bt+REEf|RBff
6 2 elexi EV
7 6 elintab Ef|RBfffRBffEf
8 imaundi RBf=RBRf
9 8 equncomi RBf=RfRB
10 9 sseq1i RBffRfRBf
11 unss RffRBfRfRBf
12 10 11 bitr4i RBffRffRBf
13 df-he RhereditaryfRff
14 13 bicomi RffRhereditaryf
15 dfss2 RBfaaRBaf
16 1 elexi BV
17 vex aV
18 16 17 elimasn aRBBaR
19 df-br BRaBaR
20 18 19 bitr4i aRBBRa
21 20 imbi1i aRBafBRaaf
22 21 albii aaRBafaBRaaf
23 15 22 bitri RBfaBRaaf
24 14 23 anbi12i RffRBfRhereditaryfaBRaaf
25 12 24 bitri RBffRhereditaryfaBRaaf
26 25 imbi1i RBffEfRhereditaryfaBRaafEf
27 impexp RhereditaryfaBRaafEfRhereditaryfaBRaafEf
28 26 27 bitri RBffEfRhereditaryfaBRaafEf
29 28 albii fRBffEffRhereditaryfaBRaafEf
30 5 7 29 3bitrri fRhereditaryfaBRaafEfBt+RE