[My solutions to A^B -> C follow...

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> Points = 9 - 6 = 3.  Looks like you owe me a Bud Lite...

Well, if either A or B is in the same bit-spot as C, and the other is either
1 or 4 bits away, then it's possible to do it in 5 instructions:

        movf    A,w     ; or "rlf", "rrf", or "swapf"
        xorwf   B,w
        xorwf   C,w
        andlw   1<<bit_c        ; See note above code snippet
        xorwf   C,f

Another 5-cycle special-case variant would be if A and B are in bit 7 and
C is in bit zero

        movf    A,w
        xorwf   B,w
        rrf     C
        addlw   128     ; Bit 7 of W into C
        rlf     C       ; Ta da!

This is, of course, not a general solution but it's faster when it works.
Alternatively, my submission for the more general case would be

        movlw   1 << bit_c
        iorwf   C,f
        btfss   A
         xorwf  C,f
        btfsc   B
         xorwf  C,f

or for a somewhat broader set of cases than the 5-cycle version above (either
A or B must be 1 or 4 bits removed from bit_c, while the other can be any bit)

        movf    A,w
        btfsc   B,bit_b
         xorlw  1 << bit_c
        xorwf   C,w
        andlw   1 << bit_c
        xorwf   C,f

This version is no shorter than the general version above, but it has the
advantage that the output will be glitch-free (it's still important that
bit C not change during the algorithm though other bits of that byte will
not be corrupted).