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No, your assumption is not correct! The language <code>L = { x = y + z | where x, y, z are binary integers and x is the sum of y and z}</code> is not Context Free Languages(CFL). I try to explain. First of all, consider following examples strings <code>s</code> in language <code>L</code>. <pre><code>110 = 100 + 10 1110 = 1100 + 10 : 111000 = 110000 + 1000 </code></pre> In my explanation LHS is <code>X</code> in question and RHS is <code>Y + Z</code>. <a href="http://en.wikipedia.org/wiki/Pumping_lemma_for_context-free_languages" rel="nofollow">What is pumping Lemma for CFL?</a> If a language L is context-free, then there exists some integer p ≥ 1 such that any string s in L with |s| ≥ p. (where p is a "pumping length" can be written as <pre><code>s = uvxyz with substrings u, v, x, y and z, such that 1. |vxy| ≤ p, 2. |vy| ≥ 1, and 3. uv nxy nz is in L for every natural number n. </code></pre> This definition | ≥ 1, and 3. uv nxy nz is in L for every natural number n. Notice: Middle part of <code>s</code> , <code>vxy</code> not greater then pumping length <code>p</code>. (condition 1) [SOLUTION]: Let us choose a string <code>s</code> in <code>L</code> that satisfy condition <code>|s| ≥ p</code> our <code>s</code> is 1m0q = 1m-10q + 10q , where <code>q > p , m-1 > p</code> Now total length of <code>s</code> is <code>2m + 2q -1</code> that is greater then <code>p</code> and of-course for some combination of natural numbers this inequality is possible (I am not including length of <code>+</code> and <code>=</code> to keep explanation simple ) Now our <code>s</code> is in language and sufficiently large according to pumping lemma for CFG. Now break it: u vxy z = 1m0q = 1m-10q + 10q Try to find <code>v</code> and <code>y</code> to pump and generate new string in language L, But keep in mind <code>v</code> and <code>y</code> should not be too much far than <code>p</code> (according to condition 1). You don't have any choice for <code>v</code> and <code>y</code> such that you can generate new strings in language! (Step-1): Because if you chose to increase <code>1</code> then you can't pump both side RHS and LHS of <code>=</code> because last <code>1</code> on LHS is at <code>q</code> (>p) to first <code>1</code> of RHS. hence not possible to generate new strings in language. (Step-2): Suppose you like to pump <code>0</code> again its not possible to increase <code>0</code> on LHS and RHS together because last <code>0</code> on LHS in m-1 (>p) distend to first <code>0</code> on RHS. (Step-3): You can't pump a combination of <code>111...000...</code> both side. , try this you will get string out of language L. Try other options too within the rules of Pumping Lemma. you would not find correct choice for <code>v</code> and <code>y</code>. [ANSWER] So he have a string <code>s</code> in L that is sufficiently large and using that we can't generate new strings in language. its contradict to Pumping Lemma for CFL hence given <code>L</code> is not a CFL. 
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1. POverifier of addition... not-regular, but is it context-free?
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