Proof: Minimum Index One
Let's prove the following theorem:
index of the mininum value in stack [ x, [ ] ] = [ 0, [ ] ]
Proof:
# | Claim | Reason |
---|---|---|
1 | index of the mininum value in stack [ x, [ ] ] = index of value (minimum value of stack [ x, [ ] ]) in [ x, [ ] ] | index of the mininum value in stack [ x, [ ] ] = index of value (minimum value of stack [ x, [ ] ]) in [ x, [ ] ] |
2 | minimum value of stack [ x, [ ] ] = x | minimum value of stack [ x, [ ] ] = x |
3 | index of value (minimum value of stack [ x, [ ] ]) in [ x, [ ] ] = index of value x in [ x, [ ] ] | if minimum value of stack [ x, [ ] ] = x, then index of value (minimum value of stack [ x, [ ] ]) in [ x, [ ] ] = index of value x in [ x, [ ] ] |
4 | index of value x in [ x, [ ] ] = index of value x in [ x, [ ] ] with current index [ 0, [ ] ] | index of value x in [ x, [ ] ] = index of value x in [ x, [ ] ] with current index [ 0, [ ] ] |
5 | x = x | x = x |
6 | index of value x in [ x, [ ] ] with current index [ 0, [ ] ] = [ 0, [ ] ] | if x = x, then index of value x in [ x, [ ] ] with current index [ 0, [ ] ] = [ 0, [ ] ] |
7 | index of value x in [ x, [ ] ] = [ 0, [ ] ] | if index of value x in [ x, [ ] ] = index of value x in [ x, [ ] ] with current index [ 0, [ ] ] and index of value x in [ x, [ ] ] with current index [ 0, [ ] ] = [ 0, [ ] ], then index of value x in [ x, [ ] ] = [ 0, [ ] ] |
8 | index of value (minimum value of stack [ x, [ ] ]) in [ x, [ ] ] = [ 0, [ ] ] | if index of value (minimum value of stack [ x, [ ] ]) in [ x, [ ] ] = index of value x in [ x, [ ] ] and index of value x in [ x, [ ] ] = [ 0, [ ] ], then index of value (minimum value of stack [ x, [ ] ]) in [ x, [ ] ] = [ 0, [ ] ] |
9 | index of the mininum value in stack [ x, [ ] ] = [ 0, [ ] ] | if index of the mininum value in stack [ x, [ ] ] = index of value (minimum value of stack [ x, [ ] ]) in [ x, [ ] ] and index of value (minimum value of stack [ x, [ ] ]) in [ x, [ ] ] = [ 0, [ ] ], then index of the mininum value in stack [ x, [ ] ] = [ 0, [ ] ] |
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