Metamath Proof Explorer


Theorem preduz

Description: The value of the predecessor class over an upper integer set. (Contributed by Scott Fenton, 16-May-2014)

Ref Expression
Assertion preduz ⊢ N ∈ ℤ ≥ M → Pred < ℤ ≥ M N = M … N − 1

Proof

Step Hyp Ref Expression
1 vex ⊢ x ∈ V
2 1 elpred ⊢ N ∈ ℤ ≥ M → x ∈ Pred < ℤ ≥ M N ↔ x ∈ ℤ ≥ M ∧ x < N
3 eluzelz ⊢ x ∈ ℤ ≥ M → x ∈ ℤ
4 eluzelz ⊢ N ∈ ℤ ≥ M → N ∈ ℤ
5 zltlem1 ⊢ x ∈ ℤ ∧ N ∈ ℤ → x < N ↔ x ≤ N − 1
6 3 4 5 syl2anr ⊢ N ∈ ℤ ≥ M ∧ x ∈ ℤ ≥ M → x < N ↔ x ≤ N − 1
7 6 pm5.32da ⊢ N ∈ ℤ ≥ M → x ∈ ℤ ≥ M ∧ x < N ↔ x ∈ ℤ ≥ M ∧ x ≤ N − 1
8 eluzel2 ⊢ N ∈ ℤ ≥ M → M ∈ ℤ
9 eluz1 ⊢ M ∈ ℤ → x ∈ ℤ ≥ M ↔ x ∈ ℤ ∧ M ≤ x
10 8 9 syl ⊢ N ∈ ℤ ≥ M → x ∈ ℤ ≥ M ↔ x ∈ ℤ ∧ M ≤ x
11 10 anbi1d ⊢ N ∈ ℤ ≥ M → x ∈ ℤ ≥ M ∧ x ≤ N − 1 ↔ x ∈ ℤ ∧ M ≤ x ∧ x ≤ N − 1
12 7 11 bitrd ⊢ N ∈ ℤ ≥ M → x ∈ ℤ ≥ M ∧ x < N ↔ x ∈ ℤ ∧ M ≤ x ∧ x ≤ N − 1
13 2 12 bitrd ⊢ N ∈ ℤ ≥ M → x ∈ Pred < ℤ ≥ M N ↔ x ∈ ℤ ∧ M ≤ x ∧ x ≤ N − 1
14 peano2zm ⊢ N ∈ ℤ → N − 1 ∈ ℤ
15 4 14 syl ⊢ N ∈ ℤ ≥ M → N − 1 ∈ ℤ
16 8 15 jca ⊢ N ∈ ℤ ≥ M → M ∈ ℤ ∧ N − 1 ∈ ℤ
17 16 biantrurd ⊢ N ∈ ℤ ≥ M → x ∈ ℤ ∧ M ≤ x ∧ x ≤ N − 1 ↔ M ∈ ℤ ∧ N − 1 ∈ ℤ ∧ x ∈ ℤ ∧ M ≤ x ∧ x ≤ N − 1
18 13 17 bitrd ⊢ N ∈ ℤ ≥ M → x ∈ Pred < ℤ ≥ M N ↔ M ∈ ℤ ∧ N − 1 ∈ ℤ ∧ x ∈ ℤ ∧ M ≤ x ∧ x ≤ N − 1
19 elfz2 ⊢ x ∈ M … N − 1 ↔ M ∈ ℤ ∧ N − 1 ∈ ℤ ∧ x ∈ ℤ ∧ M ≤ x ∧ x ≤ N − 1
20 df-3an ⊢ M ∈ ℤ ∧ N − 1 ∈ ℤ ∧ x ∈ ℤ ↔ M ∈ ℤ ∧ N − 1 ∈ ℤ ∧ x ∈ ℤ
21 20 anbi1i ⊢ M ∈ ℤ ∧ N − 1 ∈ ℤ ∧ x ∈ ℤ ∧ M ≤ x ∧ x ≤ N − 1 ↔ M ∈ ℤ ∧ N − 1 ∈ ℤ ∧ x ∈ ℤ ∧ M ≤ x ∧ x ≤ N − 1
22 anass ⊢ M ∈ ℤ ∧ N − 1 ∈ ℤ ∧ x ∈ ℤ ∧ M ≤ x ∧ x ≤ N − 1 ↔ M ∈ ℤ ∧ N − 1 ∈ ℤ ∧ x ∈ ℤ ∧ M ≤ x ∧ x ≤ N − 1
23 anass ⊢ x ∈ ℤ ∧ M ≤ x ∧ x ≤ N − 1 ↔ x ∈ ℤ ∧ M ≤ x ∧ x ≤ N − 1
24 23 anbi2i ⊢ M ∈ ℤ ∧ N − 1 ∈ ℤ ∧ x ∈ ℤ ∧ M ≤ x ∧ x ≤ N − 1 ↔ M ∈ ℤ ∧ N − 1 ∈ ℤ ∧ x ∈ ℤ ∧ M ≤ x ∧ x ≤ N − 1
25 22 24 bitr4i ⊢ M ∈ ℤ ∧ N − 1 ∈ ℤ ∧ x ∈ ℤ ∧ M ≤ x ∧ x ≤ N − 1 ↔ M ∈ ℤ ∧ N − 1 ∈ ℤ ∧ x ∈ ℤ ∧ M ≤ x ∧ x ≤ N − 1
26 19 21 25 3bitri ⊢ x ∈ M … N − 1 ↔ M ∈ ℤ ∧ N − 1 ∈ ℤ ∧ x ∈ ℤ ∧ M ≤ x ∧ x ≤ N − 1
27 18 26 bitr4di ⊢ N ∈ ℤ ≥ M → x ∈ Pred < ℤ ≥ M N ↔ x ∈ M … N − 1
28 27 eqrdv ⊢ N ∈ ℤ ≥ M → Pred < ℤ ≥ M N = M … N − 1