Proof: Reverse List Two Example

Let's prove the following theorem:

Proof:

Proof Table

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

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