Proof: Get Begin Expr Parent 12

Let's prove the following theorem:

if the following are true:

expression state at time 12 = "begin_expr"
the expression at time 12 = Dog()
parent stack at time 12 = [ ]

then parent stack at time 13 = [ Dog(), [ ] ]

Proof:

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Given

1	expression state at time 12 = "begin_expr"
2	the expression at time 12 = `Dog()`
3	parent stack at time 12 = [ ]

Proof Table

#	Claim	Reason
1	parent stack at time (12 + 1) = [ `Dog()`, parent stack at time 12 ]	if expression state at time 12 = "begin_expr" and the expression at time 12 = `Dog()`, then parent stack at time (12 + 1) = [ `Dog()`, parent stack at time 12 ] (Begin Constructor Parent Property)
2	[ `Dog()`, parent stack at time 12 ] = [ `Dog()`, [ ] ]	if parent stack at time 12 = [ ], then [ `Dog()`, parent stack at time 12 ] = [ `Dog()`, [ ] ] (Substitution)
3	parent stack at time (12 + 1) = [ `Dog()`, [ ] ]	if parent stack at time (12 + 1) = [ `Dog()`, parent stack at time 12 ] and [ `Dog()`, parent stack at time 12 ] = [ `Dog()`, [ ] ], then parent stack at time (12 + 1) = [ `Dog()`, [ ] ] (Transitive Property of Equality)
4	12 + 1 = 13	12 + 1 = 13 (Fixed Value Statement)
5	parent stack at time (12 + 1) = parent stack at time 13	if 12 + 1 = 13, then parent stack at time (12 + 1) = parent stack at time 13 (Substitution)
6	parent stack at time 13 = [ `Dog()`, [ ] ]	if parent stack at time (12 + 1) = parent stack at time 13 and parent stack at time (12 + 1) = [ `Dog()`, [ ] ], then parent stack at time 13 = [ `Dog()`, [ ] ] (Transitive Property of Equality Variation 2)

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