Proof: Maximum Index Example

Let's prove the following theorem:

In this example, we prove that the maximum index of

[3,[2,[1,[]]]]

is 0.

First, we prove that the maximum value of

[3,[2,[1,[]]]]

is 3. Then we find the index of 3 in the same list. 3 is in index 0.

Proof:

Proof Table

#	Claim	Reason
1	index of the maximum value in stack [ 3, [ 2, [ 1, [ ] ] ] ] = index of value (maximum value in stack [ 3, [ 2, [ 1, [ ] ] ] ]) in [ 3, [ 2, [ 1, [ ] ] ] ]	index of the maximum value in stack [ 3, [ 2, [ 1, [ ] ] ] ] = index of value (maximum value in stack [ 3, [ 2, [ 1, [ ] ] ] ]) in [ 3, [ 2, [ 1, [ ] ] ] ] (Maximum Index Of A List)
2	maximum value in stack [ 3, [ 2, [ 1, [ ] ] ] ] = 3	maximum value in stack [ 3, [ 2, [ 1, [ ] ] ] ] = 3 (maximum_example_2)
3	index of value (maximum value in stack [ 3, [ 2, [ 1, [ ] ] ] ]) in [ 3, [ 2, [ 1, [ ] ] ] ] = index of value 3 in [ 3, [ 2, [ 1, [ ] ] ] ]	if maximum value in stack [ 3, [ 2, [ 1, [ ] ] ] ] = 3, then index of value (maximum value in stack [ 3, [ 2, [ 1, [ ] ] ] ]) in [ 3, [ 2, [ 1, [ ] ] ] ] = index of value 3 in [ 3, [ 2, [ 1, [ ] ] ] ] (Substitution)
4	index of value 3 in [ 3, [ 2, [ 1, [ ] ] ] ] = 0	index of value 3 in [ 3, [ 2, [ 1, [ ] ] ] ] = 0 (find_number_example_2)
5	index of the maximum value in stack [ 3, [ 2, [ 1, [ ] ] ] ] = 0	if index of the maximum value in stack [ 3, [ 2, [ 1, [ ] ] ] ] = index of value (maximum value in stack [ 3, [ 2, [ 1, [ ] ] ] ]) in [ 3, [ 2, [ 1, [ ] ] ] ] and index of value (maximum value in stack [ 3, [ 2, [ 1, [ ] ] ] ]) in [ 3, [ 2, [ 1, [ ] ] ] ] = index of value 3 in [ 3, [ 2, [ 1, [ ] ] ] ] and index of value 3 in [ 3, [ 2, [ 1, [ ] ] ] ] = 0, then index of the maximum value in stack [ 3, [ 2, [ 1, [ ] ] ] ] = 0 (Transitive Property Application 1)

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