Proof: Minimum Index One

Let's prove the following theorem:

index of the mininum value in stack [ x, [ ] ] = [ 0, [ ] ]

Proof:

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Proof Table
# 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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