Requiring comprehension of equivalence to enable removal

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@garynewport , you can create "minimal working example"s of sections of your legacy code and try them out to understand the behavior per FORTRAN 77 and other dialects/extensions upon which such legacy code may be based. So for example,

! legacy
   integer, parameter :: mj = 2
   double precision x(mj,4)
   double precision, dimension(mj) :: temp, xl, r, pres
   equivalence     (x(1, 1), pres), (x(1, 4), temp)
   equivalence     (x(1, 2), r), (x(1, 3), xl)
   integer :: i
   do i = 1, mj
      x(i,1) = 1D5*i
      x(i,2) = 1D0*i
      x(i,3) = 1D1*i
      x(i,4) = 300D0*i
   end do
   print *, "pres = ", pres
   print *, "r = ", r
   print *, "x1 = ", xl
   print *, "temp = ", temp
end

vis-a-vis

! EQUIVALENCE replaced by TARGET<->POINTER paired semantics starting Fortran 90 standard
   integer, parameter :: mj = 2
   double precision, target :: x(mj,4)
   double precision, pointer, dimension(:) :: temp, xl, r, pres
   integer :: i
   pres(1:mj) => x(1:mj,1)
   r(1:mj) => x(1:mj,2)
   xl(1:mj) => x(1:mj,3)
   temp(1:mj) => x(1:mj,4)
   do i = 1, mj
      x(i,1) = 1D5*i
      x(i,2) = 1D0*i
      x(i,3) = 1D1*i
      x(i,4) = 300D0*i
   end do
   print *, "pres = ", pres
   print *, "r = ", r
   print *, "x1 = ", xl
   print *, "temp = ", temp
end

You can then run both the programs and check whether the output is equivalent - it should be the same.