It seems as though I was a bit obscure. Thus I will provide a line by line translation.
Firstly, one must understand that my husbands initials are DRH, but usually just goes by DH, dropping middle initial. My initials are now LRH, but I have hitherto been known as LR. Thus H is husbands last name.
Secondly, (<3) is a heart (in parentheses)! Gchat has taught us these essential techniques for expressing one's self.
---if(dh + lr = <3, lr=lrh, lr=lr); ---
If statement: DH plus LR equals love. If true, then LR becomes redefined as LRH. If false, LR remains LR. (LR representing the hitherto single entity). This is the IF format used in EXCEL.
---elseif(dh +lr =2*(<3), dh=dr,dh=dh);---
If statement: DH plus LR equals twice love. If true, then DH changes name to DR (takes LR’s last name), If false, DH remains DH. ELSEIF is a command in matlab I think, but it is pointless in my opinion because IF does the same job.
DH redefined to be DRH. This is just to create the new variable DRH for future use. Also note the semicolon at the end of the lines to this point. MATLAB and MAPLE lines end in this, though MAPLE also uses a colon.
---factor(drh + lrh) = (d + l)rh---
LRH has been created as a variable because the initial IF statement turned out to be true. Thus now there is both DRH and LRH variables in the workspace. Now I take the twisted turn of ignoring previously defined variables and assuming d,r,h,and l are all unique variables as though in an algebraic equation. I factor the common RH out of the equation. FACTOR is something I used to use in MAPLE and MATHEMATICA. I stopped using semicolons to terminate the lines at this point because I'm now looking at it as an algebraic equation and not a statement.
Returning to the original IF statement, since LR became LRH, DH plus LR must equal love. And since I factored DH plus LR to equal (d+l)rh, I plugged that in the love equation.
And now I am just manipulating the equation with standard mathematical operations.
---...(l-d) +d = (<3)/rh-d ---
---...(l-d) = (<3)/rh-2d ---
Aha. And now I have solved for (l-d).
---if(l - d = h) ---
I have subtracted the 12th letter of the alphabet from the 4th letter of the alphabet. This resulted in the 8th letter of the alphabet, h.
---thus: h=(<3)/rh-2d ---
I solved the above love equation for (l-d). Since I determined (l-d) equals h, I substituted h for (l-d).
---...h = (<3-2drh)/rh ---
---...rh^2 = (<3) - 2drh ---
---...rh(h+2d) = (<3) ---
This is the final line of my proof. I have redefined love (<3).
I tried and failed to find "therefore" in the symbol options or in the ALT secret combinations. Does anybody know how to make "therefore"? Alas. I know not.
---A: rh(h+2d) ---
---Q: What is love---
In Jeopardy format.