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summary of identities , for the venturesome
3.1 Exercises Motivating our First Algebraic Conjecture
3.2 Bo---ring. Isn't there a Shortcut?
3.3 Proof of Tromp
3.4 Proof of Exile/Isle of Silence
3.5 A Proven Consequence Says More
3.6 What Algebra is All About
3.7 Counting Cattle
3.8 Zero-th Law of Algebra
3.9 Algebraic Consequence of Extra Vizzo Good Luck and the Zero-th Law
3.10 Algebraic Consequence of the Pancake Operator
3.11 Another Way to Prove Algebra: The Bridge Theorem
3.12 Proof of Red Herring and School
3.13 Proof of Infect/Outfect by the Bridge Theorem
3.14 Proof of Deep Infect/Outfect by the Bridge Theorem
3.1
Exercises to Motivate First Algebraic Conjecture
Both for the sake of familiarity with the arithmetic and to get us into an algebraic issue, here are some pancake expressions for you to evaluate.
A justification will be noted for each step.
3.1.1
_____ __ ___
______ _______
__ _______________
, a given expression in side-view
( ) ((( ))
(( ) ( )))
, same expr in standard pancake
=
___ ___
_______
__ ________________
( ) (
(( ) ( )))
,
SSglax "Revoco!" spell
=
___
______
__ ______________
( ) (
( ( )))
,
EVglax "Revoco!" spell
=
__ ______________
( ) (
)
,
SSglax "Revoco!" spell
=
__
( )
,
EVglax "Revoco!" spell
simple vizzo.
3.1.2
__ __ __
________ __
____________ __ __
, a given expression in side-view
((( ) ( ) ( )) ( )) ( ) ( )
, same expr in standard pancake
=
__
________ __
____________ __ __
((( )
) ( )) ( ) ( )
,
EVglax "Revoco!" spell twice
=
__
______ __
__________ __
((( ) ) ( )) ( )
,
EVglax "Revoco!" spell
=
__
____________ __
(
( )) ( )
,
SSglax "Revoco!" spell
=
__
( ) , SSglax "Revoco!" spell ; simple vizzo.
3.1.3
__
___
__
____ __
___ ___
________ __ ________
, a given expression, side-view
(((( ))) ( )) ( ) (( )
(( )))
, same expr in standard pancake
=
____ __
___
________ __ ________
(( ) ( )) ( )
(( ) )
,
SSglax "Revoco!" spell 2 times
=
____ __
________ __
(( ( )
) ) ( )
,
SSglax "Revoco!" spell
=
____
________ __
((
) ) ( )
,
EVglax "Revoco!" spell
=
__
( ) , SSglax "Revoco!" spell ; simple vizzo.
3.1.4
__
___
__ __
___
____
___ ___ ___
__
_________ __ _____________
, given expression, side-view
( )
(( ) ((( )))
( ) ((( )) (( )) ( ))
, same expression, standard
=
___ ___
___
__ ________ __ _____________
( )
(( ) ( ))
( ) (
( ))
,
SSglax "Revoco!" 3 times
=
___ ___
__ ________ __
( ) (( ) ( ))
( )
,
SSglax "Revoco!" spell
=
___
__ ________ __
( ) (
( )) ( )
,
EVglax "Revoco!" spell
=
__ __
( ) ( )
,
SSglax "Revoco!" spell
=
__
( ) , EVglax "Revoco!" spell ; simple vizzo.
These evaluations should have become boring by now. They all came out simple vizzo, for one thing. Why?
Maybe we can use this observation to discover a shortcut. Maybe you have noticed what is causing them all to evaluate to simple vizzo.
Let's look at these given expressions to see what we might notice that they have in common:
3.1.1 ( ) ((( )) (( ) ( ))) = ( )
3.1.2 ((( ) ( ) ( )) ( )) ( ) ( ) = ( )
3.1.3 (((( ))) ( )) ( ) (( ) (( ))) = ( )
3.1.4 ( ) (( ) ((( ))) ( ) ((( )) (( )) ( )) = ( )
Look, in each example, at the unrepresented plaza. Do you notice that in every one of the four examples there is at least one empty pancake --one vizzo-- directly on the unrepresented plaza?
Is that the reason these examples all come out vizzo? If you say yes, you have discovered a powerful shortcut. As soon as you spot an empty pancake you know you can eliminate everything else on that plaza from consideration.
Can you prove this shortcut always works?
Let p represent any pancake expression. We seek to prove whether
p ( ) = ( )
in all cases (that is, in general).
This is our algebraic conjecture. Let's be bold and give it a name before we prove it.
In naming this conjecture, we might note that the vizzo seems to trump or tromp on any value that might turn out from p. So I have named it trump for the summary of pancake identities.
Now we seek to prove whether
p ( ) = ( ).
Well, how many cases are there of evaluating the subexpression p? Just two, so it's not difficult to check all cases. Start by supposing p takes the value , simple invizzo:
Case: p =
In this case, substituting invizzo for p in the proposed p ( ) = ( ) ,
we get
( ) = ( ) , substituting case p= .The proposed consequence holds in case p= .
As another case, suppose p takes the value ( ), simple vizzo:
Case: p = ( )
In this case, substitute the value vizzo for p in the proposed p ( ) = ( ) ,
and we get
( ) ( ) = ( ) , substituting p=( ).The Lhs (left-hand side) readily evaluates to vizzo:
( ) = ( ) , EVglax "Revoco!" spell
The proposed consequence holds also in case p=( ).
There are no other cases, so we say that since the Lhs evaluates to simple vizzo in all cases, the trump shortcut works:
p ( ) = ( )
trump []