Maybe you don't like talking too much about yourself But you shoulda told me that you were thinkin' 'bout someone else
You're drunk at a party or maybe it's just that your car broke down
Your phone's been off for a couple months, so you're calling me now
I know you, you're like this
When sh7t don't go your way you needed me to fix it
And like me, I did
But I ran out of every reason
Now suddenly you're asking for it back
Could you tell me, where'd you get the nerve?
Yeah, you could say you miss all that we had
But I don't really care how bad it hurts
When you broke me first
You broke me first
Took a while, was in denial when I first heard
That you moved on quicker than I could've ever, you know that hurt
Swear for a while I would stare at my phone just to see your name
But now that it's there, I don't really know what to say
I know you, you're like this
When sh7t don't go your way you needed me to fix it
And like me, I did
But I ran out of every reason
Now suddenly you're asking for it back
Could you tell me, where'd you get the nerve?
Yeah, you could say you miss all that we had
But I don't really care how bad it hurts
When you broke me first
You broke me first
What did you think would happen?
What did you think would happen?
I'll never let you have it
What did you think would happen?
Now suddenly you're asking for it back
Could you tell me, where'd you get the nerve?
Yeah, you could say you miss all that we had
But I don't really care how bad it hurts
When you broke me first
You broke me first

Step-by-step explanation:

:0

wow did you write all this?

this is good!

Step-by-step explanation:

include a picture of the drawing in order for me to

step-by-step explanation:

i think it's eleven too, don't woosh me here, but according to the "experts" it's 2.

the proof starts from the peano postulates, which define the natural

numbers n. n is the smallest set satisfying these postulates:

  p1.   1 is in n.

  p2.   if x is in n, then its "successor" x' is in n.

  p3.   there is no x such that x' = 1.

  p4.   if x isn't 1, then there is a y in n such that y' = x.

  p5.   if s is a subset of n, 1 is in s, and the implication

      (x in s => x' in s) holds, then s = n.

then you have to define addition recursively:

  def: let a and b be in n. if b = 1, then define a + b = a'

      (using p1 and p2). if b isn't 1, then let c' = b, with c in n

      (using p4), and define a + b = (a + c)'.

then you have to define 2:

  def:   2 = 1'

2 is in n by p1, p2, and the definition of 2.

theorem:   1 + 1 = 2

proof: use the first part of the definition of + with a = b = 1.

      then 1 + 1 = 1' = 2   q.e.d.

note: there is an alternate formulation of the peano postulates which

replaces 1 with 0 in p1, p3, p4, and p5. then you have to change the

definition of addition to this:

  def: let a and b be in n. if b = 0, then define a + b = a.

      if b isn't 0, then let c' = b, with c in n, and define

      a + b = (a + c)'.

you also have to define 1 = 0', and 2 = 1'. then the proof of the

theorem above is a little different:

proof: use the second part of the definition of + first:

      1 + 1 = (1 + 0)'

      now use the first part of the definition of + on the sum in

      parentheses:   1 + 1 = (1)' = 1' = 2   q.e.d.