Metamath Proof Explorer


Theorem flval3

Description: An alternate way to define the floor (greatest integer) function, as the supremum of all integers less than or equal to its argument. (Contributed by NM, 15-Nov-2004) (Proof shortened by Mario Carneiro, 6-Sep-2014)

Ref Expression
Assertion flval3 A A = sup x | x A <

Proof

Step Hyp Ref Expression
1 ssrab2 x | x A
2 zssre
3 1 2 sstri x | x A
4 3 a1i A x | x A
5 breq1 x = A x A A A
6 flcl A A
7 flle A A A
8 5 6 7 elrabd A A x | x A
9 8 ne0d A x | x A
10 reflcl A A
11 breq1 x = z x A z A
12 11 elrab z x | x A z z A
13 flge A z z A z A
14 13 biimpd A z z A z A
15 14 expimpd A z z A z A
16 12 15 syl5bi A z x | x A z A
17 16 ralrimiv A z x | x A z A
18 brralrspcev A z x | x A z A y z x | x A z y
19 10 17 18 syl2anc A y z x | x A z y
20 4 9 19 8 suprubd A A sup x | x A <
21 suprleub x | x A x | x A y z x | x A z y A sup x | x A < A z x | x A z A
22 4 9 19 10 21 syl31anc A sup x | x A < A z x | x A z A
23 17 22 mpbird A sup x | x A < A
24 4 9 19 suprcld A sup x | x A <
25 10 24 letri3d A A = sup x | x A < A sup x | x A < sup x | x A < A
26 20 23 25 mpbir2and A A = sup x | x A <