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How do I proof this theorem?? (x)(y)(z)((x=y&y=z)→x=z)

I have tried using RAA at first ~(x)(y)(z)((x=y&y=z)→x=z)

and I got ~(a=b&b=c)→a=c) after using QN and EI

then I used MI to get ~(~(a=b&b=c)v a=c)

Then I used DeM to get (~~(a=b&b=c) & ~(a=c)

At the end I have a=b, b=c and ~(a=c)

I think I need to proof that a=c to complete RAA but I don't know how to or what law to use.

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It can be proven with substitution, also known as identity elimination (=E):

{1}      1.  a=b & b=c                      Assum.
{1}      2.  a=b                            1 &E
{1}      3.  b=c                            1 &E
{1}      4.  a=c                            2,3 =E
-        5.  (a=b & b=c) → a=c              1,4 CP
-        6.  ∀z[(a=b & b=z) → a=z]          5 UI
-        7.  ∀y,z[(a=y & y=z) → a=z]        6 UI
-        8.  ∀x,y,z[(x=y & y=z) → x=z]      7 UI
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