Question

Would running array_diff_assoc() twice on an array give me all non-unique entries?

$array3 = array_diff_assoc($array1, $array2);
$array4 = array_diff_assoc($array1, $array3);
var_dump($array4);

Answer 1

Given:

• A the set of entries in $array1, and
• B the set of entries in $array2,

B would be composed of:

• B’, all the entries in B that are in A, and
• B” all the entries in B that are not in A.

$array3, diff_assoc_array($array1, $array2), would be the operation A \ B, which would reduces as follows:

• (A \ B’) ∩ (A \ B”)
• (A ∩ ¬B’) ∩ A
• A ∩ ¬B’.

$array4, diff_assoc_array($array1, $array3), would be the operation A \ (A ∩ ¬B’), which reduces as follows:

• A ∩ ¬(A ∩ ¬B’)
• A ∩ (¬A ∪ B’)
• A ∩ B

Therefore yes, the final result would be the items common to both arrays.