Proof: Minimum Index Example

Let's prove the following theorem:

In this example, we prove that the minimum index of

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

is 2.

First, we prove that the minimum value of

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

is 1. Then we find the index of 1 in the same list. 1 is in index 2.

Proof:

Proof Table

#	Claim	Reason
1	index of the mininum value in stack [ 3, [ 2, [ 1, [ ] ] ] ] = index of value (minimum value of stack [ 3, [ 2, [ 1, [ ] ] ] ]) in [ 3, [ 2, [ 1, [ ] ] ] ]	index of the mininum value in stack [ 3, [ 2, [ 1, [ ] ] ] ] = index of value (minimum value of stack [ 3, [ 2, [ 1, [ ] ] ] ]) in [ 3, [ 2, [ 1, [ ] ] ] ] (Minimum Index Of A List)
2	minimum value of stack [ 3, [ 2, [ 1, [ ] ] ] ] = 1	minimum value of stack [ 3, [ 2, [ 1, [ ] ] ] ] = 1 (minimum_example)
3	index of value (minimum value of stack [ 3, [ 2, [ 1, [ ] ] ] ]) in [ 3, [ 2, [ 1, [ ] ] ] ] = index of value 1 in [ 3, [ 2, [ 1, [ ] ] ] ]	if minimum value of stack [ 3, [ 2, [ 1, [ ] ] ] ] = 1, then index of value (minimum value of stack [ 3, [ 2, [ 1, [ ] ] ] ]) in [ 3, [ 2, [ 1, [ ] ] ] ] = index of value 1 in [ 3, [ 2, [ 1, [ ] ] ] ] (Substitution)
4	index of value 1 in [ 3, [ 2, [ 1, [ ] ] ] ] = 2	index of value 1 in [ 3, [ 2, [ 1, [ ] ] ] ] = 2 (Find Value Example)
5	index of the mininum value in stack [ 3, [ 2, [ 1, [ ] ] ] ] = 2	if index of the mininum value in stack [ 3, [ 2, [ 1, [ ] ] ] ] = index of value (minimum value of stack [ 3, [ 2, [ 1, [ ] ] ] ]) in [ 3, [ 2, [ 1, [ ] ] ] ] and index of value (minimum value of stack [ 3, [ 2, [ 1, [ ] ] ] ]) in [ 3, [ 2, [ 1, [ ] ] ] ] = index of value 1 in [ 3, [ 2, [ 1, [ ] ] ] ] and index of value 1 in [ 3, [ 2, [ 1, [ ] ] ] ] = 2, then index of the mininum value in stack [ 3, [ 2, [ 1, [ ] ] ] ] = 2 (Transitive Property Application 1)

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