Proof: Get Begin Expr Value 3

Let's prove the following theorem:

if the following are true:

expression state at time 3 = "begin_expr"
the expression at time 3 = entry "ES": "Estonia"
Value Stack at time 3 = [ [ ], [ ] ]

then Value Stack at time 4 = [ [ ], [ [ ], [ ] ] ]

Proof:

View as a tree | View dependent proofs | Try proving it

Given

1	expression state at time 3 = "begin_expr"
2	the expression at time 3 = entry "ES": "Estonia"
3	Value Stack at time 3 = [ [ ], [ ] ]

Proof Table

#	Claim	Reason
1	Value Stack at time (3 + 1) = [ [ ], Value Stack at time 3 ]	if expression state at time 3 = "begin_expr" and the expression at time 3 = entry "ES": "Estonia", then Value Stack at time (3 + 1) = [ [ ], Value Stack at time 3 ] (Begin Entry Elements (1))
2	[ [ ], Value Stack at time 3 ] = [ [ ], [ [ ], [ ] ] ]	if Value Stack at time 3 = [ [ ], [ ] ], then [ [ ], Value Stack at time 3 ] = [ [ ], [ [ ], [ ] ] ] (Substitution)
3	Value Stack at time (3 + 1) = [ [ ], [ [ ], [ ] ] ]	if Value Stack at time (3 + 1) = [ [ ], Value Stack at time 3 ] and [ [ ], Value Stack at time 3 ] = [ [ ], [ [ ], [ ] ] ], then Value Stack at time (3 + 1) = [ [ ], [ [ ], [ ] ] ] (Transitive Property of Equality)
4	3 + 1 = 4	3 + 1 = 4 (Fixed Value Statement)
5	Value Stack at time (3 + 1) = Value Stack at time 4	if 3 + 1 = 4, then Value Stack at time (3 + 1) = Value Stack at time 4 (Substitution)
6	Value Stack at time 4 = [ [ ], [ [ ], [ ] ] ]	if Value Stack at time (3 + 1) = Value Stack at time 4 and Value Stack at time (3 + 1) = [ [ ], [ [ ], [ ] ] ], then Value Stack at time 4 = [ [ ], [ [ ], [ ] ] ] (Transitive Property of Equality Variation 2)

Comments

Please log in to add comments