/

(?(DEFINE)
(?<add> \s*\+\s* )
(?<eq> \s*=\s* )
# Remove all zeroes except the last one if the number is 0
(?<zero> (?:0(?=\d))*+ )
# cl: last digit of left operand being 1, cr: last digit of right operand being 1, \d(?:0|\b) check if last digit from result is 0
# there will be carry if cl and cr are set, or cl or cr are set and the last digit from result is 0
(?<carry> (?(cl)(?(cr)|\d(?:0|\b))|(?(cr)\d(?:0|\b)|(*F))) )
# add carry with l1 (current digit of left operand being 1) and r1 (current digit of right operand being 1)
# i.e. returns result of carry + l1 + r1 in Z/2Z
(?<digitadd> (?(?= (?(?=(?(l1)(?(r1)|(*F))|(?(r1)(*F))))(?&carry)|(?!(?&carry))) )1|0) )
# check for a single digit at the current offset whether the result is correct
# ro: right operand out of bounds (i.e. the current digit is at a higher offset than the size of the left operand)
# if we're out of bounds of the right operand, cr is just not set (i.e. handled as if there were leading zeroes)
(?<recursedigit>
# now, with the r and f, we can figure out r1 and cr at the current offset and also perform binary carry addition at that offset in the result
(?&add) (?&zero) (?:\d*(?:0|1(?<r1>)))? (?(ro)|(?=(?<cr>1)?))\k<r> (?&eq) \d*(?&digitadd)\k<f>\b
# iterate through the whole left operand to find the sequences (for right operand and result) of the same length as the offset of the current digit
| (?=\d* (?&add) (?&zero) (?:\k<r>(?<ro>)|\d*(?<r>\d\k<r>)) (?&eq) \d*(?<f>\d\k<f>)\b) \d(?&recursedigit)
)
# run the check, sets l1 and cl accordingly and initializes the r (right operand) and f (final result) groups to be empty
(?<checkdigit> (?:0|1(?<l1>)) (?=(?<cl>1)?) (?<r>) (?<f>) (?&recursedigit) )
# "trivial" increment of a binary number, i.e. a +1 is applied to the part of the right operand which exceeds the length of the left operand
(?<carryoverflow>
# number contains a zero, just update the part after the last zero
(?<x>\d+) 0 (?<y> \k<r> (?&eq) 0*\k<x>1 | 1(?&y)0 )
# number contains only ones, add a leading 1 and replace all the ones by zeroes
| (?<z> 1\k<r> (?&eq) 0*10 | 1(?&z)0 )
)
# ensure correct lengths of the final operand and handle right operands being longer than the left operand
(?<recurseoverflow>
# the left operand is longer than or as long as the right one. In the latter case, the final result will always be exactly one digit longer than the operands
# in the former case, if the first non-leading zero (from the left) of the left operand is at a higher or equal offset to the length of the right operand, the final result will be one digit longer than the left operand
(?&add) 0*+ (?(rlast) \k<r> (?&eq) 0*(?(ro)(?(addone)1)|1)\k<f>\b
# the right operand has a zero at the offset equal to the length of the left operand. Then just copy the leading digits to the final result
| (?:(?<remaining>\d+)(?=0\d* (?&eq) \d*(?=1)\k<f>\b)\k<r> (?&eq) (*PRUNE) 0*\k<remaining>\k<f>\b
# otherwise there will be some carry which needs to be applied before copying the leading digits to the final result
| (?&carryoverflow)\k<f>\b))
# iterate through the whole left operand to find the sequences (for right operand and result) of the same length as the left operand
| (?=\d* (?&add) 0*+ (?:\k<r>(?<ro>)|(?=(?:\d\k<r>(?&eq)(?<rlast>))?)\d*(?<r>\d\k<r>)) (?&eq) \d*(?<f>\d\k<f>)\b)
# check - only at the first non-leading zero - whether the right operand is longer than the current offset of the iteration, or just as long and having a carry (i.e. the digit at that offset in the final result is 0)
(?(nullchecked)|(?=(?<addone>(?=0)(?=(?:\d(?=\d*(?&add)\d*(?&eq)\d*(?<c>\d\k<c>)\b))+(?&add))(?<longer>(?&add)0*|\d(?&longer)\d)(\d+(?&eq)|(?&eq)\d*(?=0)\k<c>))?)(?=(?<nullchecked>0)?))
\d(?&recurseoverflow)
)
(?<s>
(?=\d) 0*? (?<arg>[01]+)? (?&add) (?=\d) 0*? (?<arg>(?(arg)(*F))[01]+)? (?&eq) (*PRUNE) \k<arg>
| (?&zero)
# traverse the digits one by one and verify the correctness of each offset individually
(?=(?<iteratedigits> (?=(?&checkdigit))\d (?:\b|(?&iteratedigits)) ))
# assert exact format here
(?=[01]+ (?&add) [01]+ (?&eq) [01]+ \b)
# remove leading zeroes and force an additional digit on the final result in case the left operand is only ones and the right operand not longer than the left
0*? (?<r>) (?<f>) (?<c>) (?=(?<addone>1+(?&add))?) (?&recurseoverflow)
# Handle 0 + x or x + 0 separately to avoid messing around in the big subpatterns
)
)
\b(?&s)\b

/

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