#Python 2, (削除) 600 (削除ここまで)(削除) 584 (削除ここまで)(削除) 397 (削除ここまで) 373 bytes

This is my own golfed reference solution. Anyone is welcome to improve it or follow its logic in their own answer as long as attribution is given.

The neat part about it is that no recursion is done, so it will never have problems with Python's recursion limit. For example, Sp's Countup Prindeal program can run indefinitely.

p=filter(len,[l.split('#')[0].split()for l in input().split('\n')]);m={};v={};i=0
while i<len(p):
 s=p[i]
 if'('in`s`:s=s[f]
 n,f=s[0],0
 if n in m:a,b,c=([s[int(g)]if g.isdigit()else g for g in t]for t in m[n]);p=[a,(b,c)]+p[i+1:];i=0;continue
 s=s[1]
 q=v.get(s,0)
 if'd'>n:m[s]=p[i+1:i+4];i+=3
 elif'i'<n:print s,'=',q
 elif'd'<n:v[s]=q+1
 elif q:v[s]-=1
 else:f=1
 i+=1

It's a program that takes in the quoted program string with newlines escaped, e.g.
'p _MyVariable_321\np screaming_hairy_armadillo'.

I took various golfing cues from Sp's and Pietu's answers. Thanks guys :)

Python 2, (削除) 600 (削除ここまで)(削除) 584 (削除ここまで)(削除) 397 (削除ここまで) 373 bytes

This is my own golfed reference solution. Anyone is welcome to improve it or follow its logic in their own answer as long as attribution is given.

The neat part about it is that no recursion is done, so it will never have problems with Python's recursion limit. For example, Sp's Countup Prindeal program can run indefinitely.

p=filter(len,[l.split('#')[0].split()for l in input().split('\n')]);m={};v={};i=0
while i<len(p):
 s=p[i]
 if'('in`s`:s=s[f]
 n,f=s[0],0
 if n in m:a,b,c=([s[int(g)]if g.isdigit()else g for g in t]for t in m[n]);p=[a,(b,c)]+p[i+1:];i=0;continue
 s=s[1]
 q=v.get(s,0)
 if'd'>n:m[s]=p[i+1:i+4];i+=3
 elif'i'<n:print s,'=',q
 elif'd'<n:v[s]=q+1
 elif q:v[s]-=1
 else:f=1
 i+=1

It's a program that takes in the quoted program string with newlines escaped, e.g.
'p _MyVariable_321\np screaming_hairy_armadillo'.

I took various golfing cues from Sp's and Pietu's answers. Thanks guys :)

#Python 2, (削除) 600 (削除ここまで) (削除) 584 (削除ここまで) (削除) 397 (削除ここまで) 373 bytes

This is my own golfed reference solution. Anyone is welcome to improve it or follow its logic in their own answer as long as attribution is given.

The neat part about it is that no recursion is done, so it will never have problems with Python's recursion limit. For example, Sp's Countup Prindeal program can run indefinitely.

p=filter(len,[l.split('#')[0].split()for l in input().split('\n')]);m={};v={};i=0
while i<len(p):
 s=p[i]
 if'('in`s`:s=s[f]
 n,f=s[0],0
 if n in m:a,b,c=([s[int(g)]if g.isdigit()else g for g in t]for t in m[n]);p=[a,(b,c)]+p[i+1:];i=0;continue
 s=s[1]
 q=v.get(s,0)
 if'd'>n:m[s]=p[i+1:i+4];i+=3
 elif'i'<n:print s,'=',q
 elif'd'<n:v[s]=q+1
 elif q:v[s]-=1
 else:f=1
 i+=1

It's a program that takes in the quoted program string with newlines escaped, e.g.
'p _MyVariable_321\np screaming_hairy_armadillo'.

I took various golfing cues from Sp's and Pietu's answers. Thanks guys :)

#Python 2, (削除) 600 (削除ここまで) (削除) 584 (削除ここまで) (削除) 397 (削除ここまで) 373 bytes

This is my own golfed reference solution. Anyone is welcome to improve it or follow its logic in their own answer as long as attribution is given.

The neat part about it is that no recursion is done, so it will never have problems with Python's recursion limit. For example, Sp's Countup Prindeal program can run indefinitely.

p=filter(len,[l.split('#')[0].split()for l in input().split('\n')]);m={};v={};i=0
while i<len(p):
 s=p[i]
 if'('in`s`:s=s[f]
 n,f=s[0],0
 if n in m:a,b,c=([s[int(g)]if g.isdigit()else g for g in t]for t in m[n]);p=[a,(b,c)]+p[i+1:];i=0;continue
 s=s[1]
 q=v.get(s,0)
 if'd'>n:m[s]=p[i+1:i+4];i+=3
 elif'i'<n:print s,'=',q
 elif'd'<n:v[s]=q+1
 elif q:v[s]-=1
 else:f=1
 i+=1

It's a program that takes in the quoted program string with newlines escaped, e.g.
'p _MyVariable_321\np screaming_hairy_armadillo'.

I took various golfing cues from Sp's and Pietu's answers. Thanks guys :)

#Python 2, (削除) 600 (削除ここまで) (削除) 584 (削除ここまで) 397 bytes

This is my own golfed reference solution. Anyone is welcome to improve it or follow its logic in their own answer as long as attribution is given.

The neat part about it is that no recursion is done, so it will never have problems with Python's recursion limit. For example, Sp's Countup Prindeal program can run indefinitely.

p=filter(len,[l.split('#')[0].split()for l in input().split('\n')]);m={};v={};i=0
while i<len(p):
 s=p[i]
 if'('in`s`:s=s[f]
 n,f=s[0],0
 if n in m:a,b,c=([s[int(g)]if g.isdigit()else g for g in t]for t in m[n]);p=[a,(b,c)]+p[i+1:];i=0;continue
 s=s[1]
 if n<'d':m[s]=p[i+1:i+4];i+=3
 elif n>'i':print s,'=',v.get(s,0)
 elif n>'d':v[s]=1+v[s]if s in v else 1
 elif v.get(s):v[s]-=1
 else:f=1
 i+=1

It's a program that takes in the quoted program string with newlines escaped, e.g.
'p _MyVariable_321\np screaming_hairy_armadillo'.

I took various golfing cues from Sp's and Pietu's answers. Thanks guys :)

#Python 2, (削除) 600 (削除ここまで) (削除) 584 (削除ここまで) (削除) 397 (削除ここまで) 373 bytes

This is my own golfed reference solution. Anyone is welcome to improve it or follow its logic in their own answer as long as attribution is given.

The neat part about it is that no recursion is done, so it will never have problems with Python's recursion limit. For example, Sp's Countup Prindeal program can run indefinitely.

p=filter(len,[l.split('#')[0].split()for l in input().split('\n')]);m={};v={};i=0
while i<len(p):
 s=p[i]
 if'('in`s`:s=s[f]
 n,f=s[0],0
 if n in m:a,b,c=([s[int(g)]if g.isdigit()else g for g in t]for t in m[n]);p=[a,(b,c)]+p[i+1:];i=0;continue
 s=s[1]
 q=v.get(s,0)
 if'd'>n:m[s]=p[i+1:i+4];i+=3
 elif'i'<n:print s,'=',q
 elif'd'<n:v[s]=q+1
 elif q:v[s]-=1
 else:f=1
 i+=1

It's a program that takes in the quoted program string with newlines escaped, e.g.
'p _MyVariable_321\np screaming_hairy_armadillo'.

I took various golfing cues from Sp's and Pietu's answers. Thanks guys :)