Let's prove the following theorem:

reverse of remaining stack [ [ 0, [ ] ], [ [ 1, [ ] ], [ ] ] ] and already reversed stack [ ] = [ [ 1, [ ] ], [ [ 0, [ ] ], [ ] ] ]

Proof Table

#	Claim	Reason
1	reverse of remaining stack [ [ 0, [ ] ], [ [ 1, [ ] ], [ ] ] ] and already reversed stack [ ] = reverse of remaining stack [ [ 1, [ ] ], [ ] ] and already reversed stack [ [ 0, [ ] ], [ ] ]	reverse of remaining stack [ [ 0, [ ] ], [ [ 1, [ ] ], [ ] ] ] and already reversed stack [ ] = reverse of remaining stack [ [ 1, [ ] ], [ ] ] and already reversed stack [ [ 0, [ ] ], [ ] ] (Reversing a List Property 3)
2	reverse of remaining stack [ [ 1, [ ] ], [ ] ] and already reversed stack [ [ 0, [ ] ], [ ] ] = [ [ 1, [ ] ], [ [ 0, [ ] ], [ ] ] ]	reverse of remaining stack [ [ 1, [ ] ], [ ] ] and already reversed stack [ [ 0, [ ] ], [ ] ] = [ [ 1, [ ] ], [ [ 0, [ ] ], [ ] ] ] (Reversing a List Property 4)
3	reverse of remaining stack [ [ 0, [ ] ], [ [ 1, [ ] ], [ ] ] ] and already reversed stack [ ] = [ [ 1, [ ] ], [ [ 0, [ ] ], [ ] ] ]	if reverse of remaining stack [ [ 0, [ ] ], [ [ 1, [ ] ], [ ] ] ] and already reversed stack [ ] = reverse of remaining stack [ [ 1, [ ] ], [ ] ] and already reversed stack [ [ 0, [ ] ], [ ] ] and reverse of remaining stack [ [ 1, [ ] ], [ ] ] and already reversed stack [ [ 0, [ ] ], [ ] ] = [ [ 1, [ ] ], [ [ 0, [ ] ], [ ] ] ], then reverse of remaining stack [ [ 0, [ ] ], [ [ 1, [ ] ], [ ] ] ] and already reversed stack [ ] = [ [ 1, [ ] ], [ [ 0, [ ] ], [ ] ] ] (Transitive Property of Equality)

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