Proof: Pop Index Example 2

Let's prove the following theorem:

Proof:

Proof Table

#	Claim	Reason
1	remaining elements after [ 3, [ 2, [ 1, [ ] ] ] ] is popped at index 0 = remaining elements after [ 3, [ 2, [ 1, [ ] ] ] ] is popped at index 0 and visited stack is [ ]	remaining elements after [ 3, [ 2, [ 1, [ ] ] ] ] is popped at index 0 = remaining elements after [ 3, [ 2, [ 1, [ ] ] ] ] is popped at index 0 and visited stack is [ ] (Begin Pop At Index Property)
2	remaining elements after [ 3, [ 2, [ 1, [ ] ] ] ] is popped at index 0 and visited stack is [ ] = [ 2, [ 1, [ ] ] ]	remaining elements after [ 3, [ 2, [ 1, [ ] ] ] ] is popped at index 0 and visited stack is [ ] = [ 2, [ 1, [ ] ] ] (pop_index_result_example_2)
3	remaining elements after [ 3, [ 2, [ 1, [ ] ] ] ] is popped at index 0 = [ 2, [ 1, [ ] ] ]	if remaining elements after [ 3, [ 2, [ 1, [ ] ] ] ] is popped at index 0 = remaining elements after [ 3, [ 2, [ 1, [ ] ] ] ] is popped at index 0 and visited stack is [ ] and remaining elements after [ 3, [ 2, [ 1, [ ] ] ] ] is popped at index 0 and visited stack is [ ] = [ 2, [ 1, [ ] ] ], then remaining elements after [ 3, [ 2, [ 1, [ ] ] ] ] is popped at index 0 = [ 2, [ 1, [ ] ] ] (Transitive Property of Equality)

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