Proof: Reverse Two
Let's prove the following theorem:
reverse of [ [ x, [ ] ], [ [ y, [ ] ], [ ] ] ] = [ [ y, [ ] ], [ [ x, [ ] ], [ ] ] ]
Proof:
| # | Claim | Reason |
|---|---|---|
| 1 | reverse of remaining stack [ [ x, [ ] ], [ [ y, [ ] ], [ ] ] ] and already reversed stack [ ] = [ [ y, [ ] ], [ [ x, [ ] ], [ ] ] ] | reverse of remaining stack [ [ x, [ ] ], [ [ y, [ ] ], [ ] ] ] and already reversed stack [ ] = [ [ y, [ ] ], [ [ x, [ ] ], [ ] ] ] |
| 2 | 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 [ ] |
| 3 | 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, [ ] ], [ ] ] ] |
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