How do you make all the integers from 0 to 13 using four 1s

You can use any mathematical operations but don't use any other numbers & use all the four 1s. Just write expressions involving only four 1s that equal the integers from 0 to 13. So you may not use operations like ^2, ^3, ^(1/2) etc. [But may use √ instead of ^(1/2)]

1x1-1x1 = 0
1x1x1x1 = 1
1x1+1x1 = 2
1x1+1+1 = 3
1+1+1+1 = 4
(1+1+1)!-1 = 5
1x(1+1+1)! = 6
1+(1+1+1)! = 7
1/.1 -1-1 = 8
11-1-1 = 9
11-1x1 = 10
11x1x1 = 11
11+1x1 = 12
11+1+1 = 13

I don't think it's possible.... I can only get to 4.

OH COME ON, I CAN USE 11? >.>

0: 1+1-1-1
1: 1+1+1-1
3: 1*(1+1+1)
4: (1+1)*(1+1)
5: (1+1+1)!-1
6: (1+1+1)!*1
7: (1+1+1)!+1
8: ?
9: 11-1-1
10: 11-(1/1)
11: 11-(1-1)
12: 11+(1/1)
13: 11+1+1

I don't think 1/.1-1-1 should count, considering the fact that it's technically 1/(0.1)-1-1
(p.s. **** factorials.)

These problems always need to be properly defined or else the options are almost limitless.
1-4 are obvious, as are 9-13 (9 = 11 - 1 - 1 etc)

5 = (1 + 1+ 1)! - 1
6 = (1 + 1+ 1)! * 1
7 = (1+1+1)! + 1
8 = (1 / .1) - (1+1)

1 - 1 + 1 - 1 =0
1*1*1*1 = 1
1/1 + 1/1 = 2
1 + 1 + 1/1 = 3
1+1+1+1 = 4
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(1 + 1 + 1)! - 1 = 5
1*(1 + 1 + 1)! = 6
1 + (1 + 1 + 1)! = 7

Edit
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1/.1 - 1 - 1 = 8
1/.1 - (1*1) = 9
1/.1 + (1 - 1) = 10
1/.1 + (1*1) = 11
1/.1 + 1 + 1 = 12
=============

I'm going to cheat a little for 13. Nobody said I couldn't use base 2. So

13 = 1101 * 1

It seems to me that is just as good as saying 11 + 1 + 1

1= 1*1*1*1
2=1*1*1+1
3=1*1+1+1
4=1+1+1+1
5=(1+1+1)!-1
6(1+1+1)!*1
7=(1+1+1)!+1
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