Metamath Proof Explorer


Theorem dmttrcl

Description: The domain of a transitive closure is the same as the domain of the original class. (Contributed by Scott Fenton, 26-Oct-2024)

Ref Expression
Assertion dmttrcl dom t++ R = dom R

Proof

Step Hyp Ref Expression
1 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
2 1 dmeqi dom t++ R = dom x y | n ω 1 𝑜 f f Fn suc n f = x f n = y a n f a R f suc a
3 dmopab dom x y | n ω 1 𝑜 f f Fn suc n f = x f n = y a n f a R f suc a = x | y n ω 1 𝑜 f f Fn suc n f = x f n = y a n f a R f suc a
4 2 3 eqtri dom 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
5 simpr2l n ω 1 𝑜 f Fn suc n f = x f n = y a n f a R f suc a f = x
6 fveq2 a = f a = f
7 suceq a = suc a = suc
8 df-1o 1 𝑜 = suc
9 7 8 eqtr4di a = suc a = 1 𝑜
10 9 fveq2d a = f suc a = f 1 𝑜
11 6 10 breq12d a = f a R f suc a f R f 1 𝑜
12 simpr3 n ω 1 𝑜 f Fn suc n f = x f n = y a n f a R f suc a a n f a R f suc a
13 eldif n ω 1 𝑜 n ω ¬ n 1 𝑜
14 0ex V
15 nnord n ω Ord n
16 ordelsuc V Ord n n suc n
17 14 15 16 sylancr n ω n suc n
18 8 sseq1i 1 𝑜 n suc n
19 1on 1 𝑜 On
20 19 onordi Ord 1 𝑜
21 ordtri1 Ord 1 𝑜 Ord n 1 𝑜 n ¬ n 1 𝑜
22 20 15 21 sylancr n ω 1 𝑜 n ¬ n 1 𝑜
23 18 22 bitr3id n ω suc n ¬ n 1 𝑜
24 17 23 bitr2d n ω ¬ n 1 𝑜 n
25 24 biimpa n ω ¬ n 1 𝑜 n
26 13 25 sylbi n ω 1 𝑜 n
27 26 adantr n ω 1 𝑜 f Fn suc n f = x f n = y a n f a R f suc a n
28 11 12 27 rspcdva n ω 1 𝑜 f Fn suc n f = x f n = y a n f a R f suc a f R f 1 𝑜
29 5 28 eqbrtrrd n ω 1 𝑜 f Fn suc n f = x f n = y a n f a R f suc a x R f 1 𝑜
30 vex x V
31 fvex f 1 𝑜 V
32 30 31 breldm x R f 1 𝑜 x dom R
33 29 32 syl n ω 1 𝑜 f Fn suc n f = x f n = y a n f a R f suc a x dom R
34 33 ex n ω 1 𝑜 f Fn suc n f = x f n = y a n f a R f suc a x dom R
35 34 exlimdv n ω 1 𝑜 f f Fn suc n f = x f n = y a n f a R f suc a x dom R
36 35 rexlimiv n ω 1 𝑜 f f Fn suc n f = x f n = y a n f a R f suc a x dom R
37 36 exlimiv y n ω 1 𝑜 f f Fn suc n f = x f n = y a n f a R f suc a x dom R
38 37 abssi x | y n ω 1 𝑜 f f Fn suc n f = x f n = y a n f a R f suc a dom R
39 4 38 eqsstri dom t++ R dom R
40 dmresv dom R V = dom R
41 relres Rel R V
42 ssttrcl Rel R V R V t++ R V
43 41 42 ax-mp R V t++ R V
44 ttrclresv t++ R V = t++ R
45 43 44 sseqtri R V t++ R
46 dmss R V t++ R dom R V dom t++ R
47 45 46 ax-mp dom R V dom t++ R
48 40 47 eqsstrri dom R dom t++ R
49 39 48 eqssi dom t++ R = dom R