Metamath Proof Explorer


Theorem bnj1127

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

Ref Expression
Assertion bnj1127 Y trCl X A R Y A

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 eqid ω = ω
4 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
5 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
6 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 f i A 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 i A
7 biid j n j E i [˙j / i]˙ 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 i A j n j E i [˙j / i]˙ 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 i A
8 biid [˙j / i]˙ f = pred X A R [˙j / i]˙ f = pred X A R
9 biid [˙j / i]˙ i ω suc i n f suc i = y f i pred y A R [˙j / i]˙ i ω suc i n f suc i = y f i pred y A R
10 biid [˙j / i]˙ 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 [˙j / i]˙ 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
11 biid [˙j / i]˙ 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 i A [˙j / i]˙ 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 i A
12 1 2 3 4 5 6 7 8 9 10 11 bnj1128 Y trCl X A R Y A