6.0
macs:Mac OS X10.4.1
SntP
20
@x (/y (x = y) $ x = d)
; Do you see what this is trying to say about the 
; world?  (Notice the names.) You can't express 
; that in the blocks language.  Negate the whole 
; sentence.@x @y ((Cube(x) & Cube(y)) $ /z Between(z, x, y))
; Now this sentence tries to make a reasonable 
; claim.  Why is it false?  Play the game and see. 
; Then fix it so it says what was intended.@x (Between(x, d, c) $ x = b)@x (/y Between(x, y, c) $ x = b)@x (/y Between(x, y, c) $ ~Large(x))@x (/y /z Between(x, y, z) $ ~Large(x))@x (/y /z Between(x, y, z) $ Tet(x))@x (~/y LeftOf(y, x) $ x = a)@x ((~/y LeftOf(y, x) & ~/y FrontOf(y, x)) $ x = a)@x (/y /z (Between(x, y, z) & Tet(y) & Tet(z)) 
                   $ x = e)@x (/y /z (Between(x, y, z) & Cube(y) & Cube(z)) 
                   $ x = b)@y (/x /z (Between(x, y, z) & x = b) $ 
                 (y = a | y = c))@x @y ((Tet(x) & Small(x) & Tet(y) & Small(y))
             $ x = y)
; Do you see what this says?@x @y ((Dodec(x) & Small(x) & Dodec(y) & Small(y))
             $ x = y)
; If you understood the last one, you see why this is
; true as well.@x (Dodec(x) $ x = b)
; This may look like a dumb thing to say, but make
; sure you understand it.  Why is it true?@x (Dodec(x) % x = b)
; And do you see why this is false?  Under what 
; circumstances would it be true?@x ((Tet(x) & Small(x)) % x = b)
; This time, we got it right./y @x ((Tet(x) & Small(x)) % x = y)
; Compare this with the previous sentence.  Do 
; you understand what it says?  Play the game 
; a couple times, committed to both true and false./y @x ((Cube(x) & Small(x)) % x = y)
; Play the game here, too.  Do you see why you can't 
; win when committed to true?
; In this slot, write a sentence that says there's 
; exactly one large tetrahedron.  Pattern it on 
; sentence 18.
s=294329;