#Python 2, 486 bytes

This is the reference solution which I golfed more (currently -98 bytes).

import sys;sys.setrecursionlimit(2000)
def r(s):
 n=s[0]
 if n in A:f=lambda i:r([s[int(t)]if'0'<t[0]<':'else t for t in A[n][i]]);return f(1+(f(0)or 0))
 k=s[1]
 if'i'<n:print k,'=',V.get(k,0)
 elif'd'<n:V[k]=-~V[k]if k in V else 1
 elif'a'<n:
 if~-(k in V)or V[k]<1:return 1
 V[k]-=1
 else:A[k]=s[2:]
A={};V={};c=filter(bool,([l,l[:l.find('#')]]['#'in l]for l in input().split('\n')))
while c:
 s=c[0].split();c=c[1:]
 if'a'!=s[0]:r(s)
 else:r(['a',s[1]]+map(str.split,c[:3]));c=c[3:]

Changes (that I recall):

• automatic boolean-integer conversion ([l,l[:l.find('#')]]['#'in l]).
• set or increment in one statement (V[k]=-~V[k]if k in V else 1)
• more aliases to longer expressions (k=s[1])
• no counter in the main loop, clearing the input list instead
• print automatically adding spaces (print k,'=',V.get(k,0))
• checking digits 1-9 ('0'<t[0]<':')
• flipping the return values of r around to save returns
• removing repetition of slicing and splitting (map(str.split,c[:3])))

Python 2, 486 bytes

This is the reference solution which I golfed more (currently -98 bytes).

import sys;sys.setrecursionlimit(2000)
def r(s):
 n=s[0]
 if n in A:f=lambda i:r([s[int(t)]if'0'<t[0]<':'else t for t in A[n][i]]);return f(1+(f(0)or 0))
 k=s[1]
 if'i'<n:print k,'=',V.get(k,0)
 elif'd'<n:V[k]=-~V[k]if k in V else 1
 elif'a'<n:
 if~-(k in V)or V[k]<1:return 1
 V[k]-=1
 else:A[k]=s[2:]
A={};V={};c=filter(bool,([l,l[:l.find('#')]]['#'in l]for l in input().split('\n')))
while c:
 s=c[0].split();c=c[1:]
 if'a'!=s[0]:r(s)
 else:r(['a',s[1]]+map(str.split,c[:3]));c=c[3:]

Changes (that I recall):

• automatic boolean-integer conversion ([l,l[:l.find('#')]]['#'in l]).
• set or increment in one statement (V[k]=-~V[k]if k in V else 1)
• more aliases to longer expressions (k=s[1])
• no counter in the main loop, clearing the input list instead
• print automatically adding spaces (print k,'=',V.get(k,0))
• checking digits 1-9 ('0'<t[0]<':')
• flipping the return values of r around to save returns
• removing repetition of slicing and splitting (map(str.split,c[:3])))

#Python 2, 486 bytes

This is the reference solution which I golfed more (currently -98 bytes).

import sys;sys.setrecursionlimit(2000)
def r(s):
 n=s[0]
 if n in A:f=lambda i:r([s[int(t)]if'0'<t[0]<':'else t for t in A[n][i]]);return f(1+(f(0)or 0))
 k=s[1]
 if'i'<n:print k,'=',V.get(k,0)
 elif'd'<n:V[k]=-~V[k]if k in V else 1
 elif'a'<n:
 if~-(k in V)or V[k]<1:return 1
 V[k]-=1
 else:A[k]=s[2:]
A={};V={};c=filter(bool,([l,l[:l.find('#')]]['#'in l]for l in input().split('\n')))
while c:
 s=c[0].split();c=c[1:]
 if'a'!=s[0]:r(s)
 else:r(['a',s[1]]+map(str.split,c[:3]));c=c[3:]

Changes (that I recall):

• automatic boolean-integer conversion ([l,l[:l.find('#')]]['#'in l]).
• set or increment in one statement (V[k]=-~V[k]if k in V else 1)
• more aliases to longer expressions (k=s[1])
• no counter in the main loop, clearing the input list instead
• print automatically adding spaces (print k,'=',V.get(k,0))
• checking digits 1-9 ('0'<t[0]<':')
• flipping the return values of r around to save returns
• removing repetition of slicing and splitting (map(str.split,c[:3])))

#Python 2, 486 bytes

This is the reference solution which I golfed more (currently -98 bytes).

import sys;sys.setrecursionlimit(2000)
def r(s):
 n=s[0]
 if n in A:f=lambda i:r([s[int(t)]if'0'<t[0]<':'else t for t in A[n][i]]);return f(1+(f(0)or 0))
 k=s[1]
 if'i'<n:print k,'=',V.get(k,0)
 elif'd'<n:V[k]=-~V[k]if k in V else 1
 elif'a'<n:
 if~-(k in V)or V[k]<1:return 1
 V[k]-=1
 else:A[k]=s[2:]
A={};V={};c=filter(bool,([l,l[:l.find('#')]]['#'in l]for l in input().split('\n')))
while c:
 s=c[0].split();c=c[1:]
 if'a'!=s[0]:r(s)
 else:r(['a',s[1]]+map(str.split,c[:3]));c=c[3:]

Changes (that I recall):

• automatic boolean-integer conversion ([l,l[:l.find('#')]]['#'in l]).
• set or increment in one statement (V[k]=-~V[k]if k in V else 1)
• more aliases to longer expressions (k=s[1])
• no counter in the main loop, clearing the input list instead
• print automatically adding spaces (print k,'=',V.get(k,0))
• checking digits 1-9 ('0'<t[0]<':')
• flipping the return values of r around to save returns
• removing repetition of slicing and splitting (map(str.split,c[:3])))