Metamath Proof Explorer


Theorem brttrcl

Description: Characterization of elements of the transitive closure of a relation. (Contributed by Scott Fenton, 18-Aug-2024)

Ref Expression
Assertion brttrcl A t++ R B n ω 1 𝑜 f f Fn suc n f = A f n = B a n f a R f suc a

Proof

Step Hyp Ref Expression
1 relttrcl Rel t++ R
2 1 brrelex12i A t++ R B A V B V
3 fvex f V
4 eleq1 f = A f V A V
5 3 4 mpbii f = A A V
6 fvex f n V
7 eleq1 f n = B f n V B V
8 6 7 mpbii f n = B B V
9 5 8 anim12i f = A f n = B A V B V
10 9 3ad2ant2 f Fn suc n f = A f n = B a n f a R f suc a A V B V
11 10 exlimiv f f Fn suc n f = A f n = B a n f a R f suc a A V B V
12 11 rexlimivw n ω 1 𝑜 f f Fn suc n f = A f n = B a n f a R f suc a A V B V
13 eqeq2 x = A f = x f = A
14 13 anbi1d x = A f = x f n = y f = A f n = y
15 14 3anbi2d x = A f Fn suc n f = x f n = y a n f a R f suc a f Fn suc n f = A f n = y a n f a R f suc a
16 15 exbidv x = A f f Fn suc n f = x f n = y a n f a R f suc a f f Fn suc n f = A f n = y a n f a R f suc a
17 16 rexbidv x = A n ω 1 𝑜 f f Fn suc n f = x f n = y a n f a R f suc a n ω 1 𝑜 f f Fn suc n f = A f n = y a n f a R f suc a
18 eqeq2 y = B f n = y f n = B
19 18 anbi2d y = B f = A f n = y f = A f n = B
20 19 3anbi2d y = B f Fn suc n f = A f n = y a n f a R f suc a f Fn suc n f = A f n = B a n f a R f suc a
21 20 exbidv y = B f f Fn suc n f = A f n = y a n f a R f suc a f f Fn suc n f = A f n = B a n f a R f suc a
22 21 rexbidv y = B n ω 1 𝑜 f f Fn suc n f = A f n = y a n f a R f suc a n ω 1 𝑜 f f Fn suc n f = A f n = B a n f a R f suc a
23 df-ttrcl t++ R = x y | n ω 1 𝑜 f f Fn suc n f = x f n = y a n f a R f suc a
24 17 22 23 brabg A V B V A t++ R B n ω 1 𝑜 f f Fn suc n f = A f n = B a n f a R f suc a
25 2 12 24 pm5.21nii A t++ R B n ω 1 𝑜 f f Fn suc n f = A f n = B a n f a R f suc a