Proof: Pop Index Result 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 and visited stack is [ ] = reverse of (result of dumping [ 2, [ 1, [ ] ] ] to [ ])	remaining elements after [ 3, [ 2, [ 1, [ ] ] ] ] is popped at index 0 and visited stack is [ ] = reverse of (result of dumping [ 2, [ 1, [ ] ] ] to [ ]) (Pop At Index Zero Property)
2	result of dumping [ 2, [ 1, [ ] ] ] to [ ] = [ 1, [ 2, [ ] ] ]	result of dumping [ 2, [ 1, [ ] ] ] to [ ] = [ 1, [ 2, [ ] ] ] (push_stack_example_2)
3	reverse of (result of dumping [ 2, [ 1, [ ] ] ] to [ ]) = reverse of [ 1, [ 2, [ ] ] ]	if result of dumping [ 2, [ 1, [ ] ] ] to [ ] = [ 1, [ 2, [ ] ] ], then reverse of (result of dumping [ 2, [ 1, [ ] ] ] to [ ]) = reverse of [ 1, [ 2, [ ] ] ] (Substitution)
4	reverse of [ 1, [ 2, [ ] ] ] = [ 2, [ 1, [ ] ] ]	reverse of [ 1, [ 2, [ ] ] ] = [ 2, [ 1, [ ] ] ] (reverse_list_two_example)
5	reverse of (result of dumping [ 2, [ 1, [ ] ] ] to [ ]) = [ 2, [ 1, [ ] ] ]	if reverse of (result of dumping [ 2, [ 1, [ ] ] ] to [ ]) = reverse of [ 1, [ 2, [ ] ] ] and reverse of [ 1, [ 2, [ ] ] ] = [ 2, [ 1, [ ] ] ], then reverse of (result of dumping [ 2, [ 1, [ ] ] ] to [ ]) = [ 2, [ 1, [ ] ] ] (Transitive Property of Equality)
6	remaining elements after [ 3, [ 2, [ 1, [ ] ] ] ] is popped at index 0 and visited stack is [ ] = [ 2, [ 1, [ ] ] ]	if remaining elements after [ 3, [ 2, [ 1, [ ] ] ] ] is popped at index 0 and visited stack is [ ] = reverse of (result of dumping [ 2, [ 1, [ ] ] ] to [ ]) and reverse of (result of dumping [ 2, [ 1, [ ] ] ] to [ ]) = [ 2, [ 1, [ ] ] ], then remaining elements after [ 3, [ 2, [ 1, [ ] ] ] ] is popped at index 0 and visited stack is [ ] = [ 2, [ 1, [ ] ] ] (Transitive Property of Equality)

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