Challenge
Create a function takes in two 2-dimensional arrays of Characters (or Strings if the programming language does not have characters as a datatype) as inputs: a and b.

Your task is to determine if b contains a. If this is so, return true. Otherwise, return false.

Sample Test Cases

a:

123
456
789
b:

123
456
789

should return true.

a:

code
golf
b:

thisis
code!!
golf!!
ohyeah

should return true.

a:

abcd
efgh
ijkl
b:

abcdef
ghijkl
mnopqr

should return false.

Least bytes wins.

Brachylog (v2), 4 bytes

ss

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Most easily run as a full program, as usual for a decision-problem, with a specified as a command-line argument, b on standard input. The question asks for a function, and the program also works as a function, with b on the left, a on the right, and output via producing an exception if and only if the decision is false.

Explanation

ss
s a substring of rows of the left input
 … assert rectangular; swap row and column operations
 s a substring of <s>rows</s> columns of the above matrix
 implicit assert that the result can be the right input

The "assert rectangular" is, obviously, pointless, as the question guarantees that already. The rest of the program does the grid-finding for us by identifying a substring of the rows and of the columns, i.e. a submatrix.

Meta-discussion

We've had a very similar question before; I'd expect most answers to one question to be modifiable into answers to the other. I think this is the neater version of it, though.

Python 2, 106 118 bytes

lambda a,b,L=len:any(any(sum(A==B[j:j+L(A)]for A,B in zip(a,b[i:]))==L(a)for i in range(L(b)))for j in range(L(b[0])))

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Python 2, 160 139 bytes

a,b=input()
l=len(b)+1
p=len(b[0])+1
print any(a==[d[s%p:n]for d in b][s/p:e]for s in range(l*p)for e in range(s/p,l)for n in range(s%p,p))

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Generates all possible and compares them with 'a'. I did not assume for it to be square, as it was not stated. But that would make the code a bit smaller.

Original code for understanding purposes:

def f(a,b):
 col_len = len(b)+1
 row_len = len(b[0])+1
 for start_col in range(col_len):
 for start_row in range(row_len):
 for end_col in range(start_col, col_len):
 for end_row in range(start_row, row_len):
 t=list(map(lambda a:a[start_row:end_row],b))[start_col:end_col]
 #print(t, a)
 if t == a:
 return True
 return False

Version that supports only squares (125 bytes):

a,b=input()
l=len(b)+1
print any(a==[d[s%l:n]for d in b][s/l:e]for s in range(l*l)for e in range(s/l,l)for n in range(s%l,l))

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J, 21 bytes

<@[(1 e.[:,=)$@[<;.3]

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how

• 
  <@[ boxed left arg


• 
  $@[<;.3] all rectangles in the right arg with the same shape as the left arg


• now pass those to things as the left and right arg to...


• 
  (1 e.[:,=) is 1 an element of e. the flatten of [:, the boolean matrix showing if the left arg equals any of the rectangles in the right arg.

python 3, 50 bytes

# assumes input formated like this
a = ["123","456","789"]
b = ["123","456","789"]

f=lambda a,b:min(max(c in d for d in b)for c in a)

edit: github with working testsuite