Metamath Proof Explorer


Theorem relexp1g

Description: A relation composed once is itself. (Contributed by RP, 22-May-2020)

Ref Expression
Assertion relexp1g ⊢ R ∈ V → R ↑ r 1 = R

Proof

Step Hyp Ref Expression
1 df-relexp ⊢ ↑ r = r ∈ V , n ∈ ℕ 0 ⟼ if n = 0 I ↾ dom ⁡ r ∪ ran ⁡ r seq 1 x ∈ V , y ∈ V ⟼ x ∘ r z ∈ V ⟼ r ⁡ n
2 1 a1i ⊢ R ∈ V → ↑ r = r ∈ V , n ∈ ℕ 0 ⟼ if n = 0 I ↾ dom ⁡ r ∪ ran ⁡ r seq 1 x ∈ V , y ∈ V ⟼ x ∘ r z ∈ V ⟼ r ⁡ n
3 simprr ⊢ R ∈ V ∧ r = R ∧ n = 1 → n = 1
4 ax-1ne0 ⊢ 1 ≠ 0
5 neeq1 ⊢ n = 1 → n ≠ 0 ↔ 1 ≠ 0
6 4 5 mpbiri ⊢ n = 1 → n ≠ 0
7 3 6 syl ⊢ R ∈ V ∧ r = R ∧ n = 1 → n ≠ 0
8 7 neneqd ⊢ R ∈ V ∧ r = R ∧ n = 1 → ¬ n = 0
9 8 iffalsed ⊢ R ∈ V ∧ r = R ∧ n = 1 → if n = 0 I ↾ dom ⁡ r ∪ ran ⁡ r seq 1 x ∈ V , y ∈ V ⟼ x ∘ r z ∈ V ⟼ r ⁡ n = seq 1 x ∈ V , y ∈ V ⟼ x ∘ r z ∈ V ⟼ r ⁡ n
10 simprl ⊢ R ∈ V ∧ r = R ∧ n = 1 → r = R
11 10 mpteq2dv ⊢ R ∈ V ∧ r = R ∧ n = 1 → z ∈ V ⟼ r = z ∈ V ⟼ R
12 11 seqeq3d ⊢ R ∈ V ∧ r = R ∧ n = 1 → seq 1 x ∈ V , y ∈ V ⟼ x ∘ r z ∈ V ⟼ r = seq 1 x ∈ V , y ∈ V ⟼ x ∘ r z ∈ V ⟼ R
13 12 3 fveq12d ⊢ R ∈ V ∧ r = R ∧ n = 1 → seq 1 x ∈ V , y ∈ V ⟼ x ∘ r z ∈ V ⟼ r ⁡ n = seq 1 x ∈ V , y ∈ V ⟼ x ∘ r z ∈ V ⟼ R ⁡ 1
14 1z ⊢ 1 ∈ ℤ
15 eqidd ⊢ R ∈ V ∧ r = R ∧ n = 1 → z ∈ V ⟼ R = z ∈ V ⟼ R
16 eqidd ⊢ R ∈ V ∧ r = R ∧ n = 1 ∧ z = 1 → R = R
17 1ex ⊢ 1 ∈ V
18 17 a1i ⊢ R ∈ V ∧ r = R ∧ n = 1 → 1 ∈ V
19 simpl ⊢ R ∈ V ∧ r = R ∧ n = 1 → R ∈ V
20 15 16 18 19 fvmptd ⊢ R ∈ V ∧ r = R ∧ n = 1 → z ∈ V ⟼ R ⁡ 1 = R
21 14 20 seq1i ⊢ R ∈ V ∧ r = R ∧ n = 1 → seq 1 x ∈ V , y ∈ V ⟼ x ∘ r z ∈ V ⟼ R ⁡ 1 = R
22 9 13 21 3eqtrd ⊢ R ∈ V ∧ r = R ∧ n = 1 → if n = 0 I ↾ dom ⁡ r ∪ ran ⁡ r seq 1 x ∈ V , y ∈ V ⟼ x ∘ r z ∈ V ⟼ r ⁡ n = R
23 elex ⊢ R ∈ V → R ∈ V
24 1nn0 ⊢ 1 ∈ ℕ 0
25 24 a1i ⊢ R ∈ V → 1 ∈ ℕ 0
26 2 22 23 25 23 ovmpod ⊢ R ∈ V → R ↑ r 1 = R