Metamath Proof Explorer


Theorem bnj1029

Description: Property of _trCl . (Contributed by Jonathan Ben-Naim, 3-Jun-2011) (New usage is discouraged.)

Ref Expression
Assertion bnj1029 R FrSe A X A TrFo trCl X A R A R

Proof

Step Hyp Ref Expression
1 biid f = pred X A R f = pred X A R
2 biid i ω suc i n f suc i = y f i pred y A R i ω suc i n f suc i = y f i pred y A R
3 biid n ω f Fn n f = pred X A R i ω suc i n f suc i = y f i pred y A R n ω f Fn n f = pred X A R i ω suc i n f suc i = y f i pred y A R
4 biid R FrSe A X A y trCl X A R z pred y A R R FrSe A X A y trCl X A R z pred y A R
5 biid m ω n = suc m p = suc n m ω n = suc m p = suc n
6 biid i n y f i i n y f i
7 biid [˙p / n]˙ f = pred X A R [˙p / n]˙ f = pred X A R
8 biid [˙p / n]˙ i ω suc i n f suc i = y f i pred y A R [˙p / n]˙ i ω suc i n f suc i = y f i pred y A R
9 biid [˙p / n]˙ n ω f Fn n f = pred X A R i ω suc i n f suc i = y f i pred y A R [˙p / n]˙ n ω f Fn n f = pred X A R i ω suc i n f suc i = y f i pred y A R
10 biid [˙ f n y f m pred y A R / f]˙ [˙p / n]˙ f = pred X A R [˙ f n y f m pred y A R / f]˙ [˙p / n]˙ f = pred X A R
11 biid [˙ f n y f m pred y A R / f]˙ [˙p / n]˙ i ω suc i n f suc i = y f i pred y A R [˙ f n y f m pred y A R / f]˙ [˙p / n]˙ i ω suc i n f suc i = y f i pred y A R
12 biid [˙ f n y f m pred y A R / f]˙ [˙p / n]˙ n ω f Fn n f = pred X A R i ω suc i n f suc i = y f i pred y A R [˙ f n y f m pred y A R / f]˙ [˙p / n]˙ n ω f Fn n f = pred X A R i ω suc i n f suc i = y f i pred y A R
13 eqid ω = ω
14 eqid f | n ω f Fn n f = pred X A R i ω suc i n f suc i = y f i pred y A R = f | n ω f Fn n f = pred X A R i ω suc i n f suc i = y f i pred y A R
15 eqid y f m pred y A R = y f m pred y A R
16 eqid f n y f m pred y A R = f n y f m pred y A R
17 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 bnj907 R FrSe A X A TrFo trCl X A R A R