Truth Propagation Property 2

Maximum Example

Find Number Example 3

Transitive Property Application 1

Maximum Index Example 2

Proof: Maximum Index Example 2

Let's prove the following theorem:

index of the maximum value in stack [ 2, [ 1, [ ] ] ] = 0

Proof:

View as a tree | View dependent proofs | Try proving it

Proof Table

#	Claim	Reason
1	index of the maximum value in stack [ 2, [ 1, [ ] ] ] = index of value (maximum value in stack [ 2, [ 1, [ ] ] ]) in [ 2, [ 1, [ ] ] ]	index of the maximum value in stack [ 2, [ 1, [ ] ] ] = index of value (maximum value in stack [ 2, [ 1, [ ] ] ]) in [ 2, [ 1, [ ] ] ] (Maximum Index Of A List)
2	maximum value in stack [ 2, [ 1, [ ] ] ] = 2	maximum value in stack [ 2, [ 1, [ ] ] ] = 2 (maximum_example)
3	index of value (maximum value in stack [ 2, [ 1, [ ] ] ]) in [ 2, [ 1, [ ] ] ] = index of value 2 in [ 2, [ 1, [ ] ] ]	if maximum value in stack [ 2, [ 1, [ ] ] ] = 2, then index of value (maximum value in stack [ 2, [ 1, [ ] ] ]) in [ 2, [ 1, [ ] ] ] = index of value 2 in [ 2, [ 1, [ ] ] ] (Substitution)
4	index of value 2 in [ 2, [ 1, [ ] ] ] = 0	index of value 2 in [ 2, [ 1, [ ] ] ] = 0 (find_number_example_3)
5	index of the maximum value in stack [ 2, [ 1, [ ] ] ] = 0	if index of the maximum value in stack [ 2, [ 1, [ ] ] ] = index of value (maximum value in stack [ 2, [ 1, [ ] ] ]) in [ 2, [ 1, [ ] ] ] and index of value (maximum value in stack [ 2, [ 1, [ ] ] ]) in [ 2, [ 1, [ ] ] ] = index of value 2 in [ 2, [ 1, [ ] ] ] and index of value 2 in [ 2, [ 1, [ ] ] ] = 0, then index of the maximum value in stack [ 2, [ 1, [ ] ] ] = 0 (Transitive Property Application 1)

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