Proof: Pop Index Example 3

Let's prove the following theorem:

Proof:

Proof Table

#	Claim	Reason
1	remaining elements after [ x, [ y, [ ] ] ] is popped at index [ 1, [ ] ] and visited stack is [ ] = remaining elements after [ y, [ ] ] is popped at index [ 0, [ ] ] and visited stack is [ x, [ ] ]	remaining elements after [ x, [ y, [ ] ] ] is popped at index [ 1, [ ] ] and visited stack is [ ] = remaining elements after [ y, [ ] ] is popped at index [ 0, [ ] ] and visited stack is [ x, [ ] ] (Pop Index (3))
2	remaining elements after [ y, [ ] ] is popped at index [ 0, [ ] ] and visited stack is [ x, [ ] ] = result of dumping [ ] to [ x, [ ] ]	remaining elements after [ y, [ ] ] is popped at index [ 0, [ ] ] and visited stack is [ x, [ ] ] = result of dumping [ ] to [ x, [ ] ] (Pop Index (2))
3	result of dumping [ ] to [ x, [ ] ] = [ x, [ ] ]	result of dumping [ ] to [ x, [ ] ] = [ x, [ ] ] (Pre-extend (3))
4	remaining elements after [ y, [ ] ] is popped at index [ 0, [ ] ] and visited stack is [ x, [ ] ] = [ x, [ ] ]	if remaining elements after [ y, [ ] ] is popped at index [ 0, [ ] ] and visited stack is [ x, [ ] ] = result of dumping [ ] to [ x, [ ] ] and result of dumping [ ] to [ x, [ ] ] = [ x, [ ] ], then remaining elements after [ y, [ ] ] is popped at index [ 0, [ ] ] and visited stack is [ x, [ ] ] = [ x, [ ] ] (Transitive Property of Equality)
5	remaining elements after [ x, [ y, [ ] ] ] is popped at index [ 1, [ ] ] and visited stack is [ ] = [ x, [ ] ]	if remaining elements after [ x, [ y, [ ] ] ] is popped at index [ 1, [ ] ] and visited stack is [ ] = remaining elements after [ y, [ ] ] is popped at index [ 0, [ ] ] and visited stack is [ x, [ ] ] and remaining elements after [ y, [ ] ] is popped at index [ 0, [ ] ] and visited stack is [ x, [ ] ] = [ x, [ ] ], then remaining elements after [ x, [ y, [ ] ] ] is popped at index [ 1, [ ] ] and visited stack is [ ] = [ x, [ ] ] (Transitive Property of Equality)

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