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Proof step 5 from proof Del-rec-other (previous step)

⊦ str = str0, (str1 ∈ sdt[str0, tree] -- str ↔ str1 ∈ (sdt -- str)[str0, tree]) ∧ (sdt[str0, tree] -- str)[str1] = (sdt -- str)[str0, tree][str1]

User applied rule case distinction

And obtained the following new goals

Proof step 6:
⊦ str1 ∈ sdt[str0, tree] -- str ↔ str1 ∈ (sdt -- str)[str0, tree], str = str0
Proof step 10:
⊦ (sdt[str0, tree] -- str)[str1] = (sdt -- str)[str0, tree][str1], str = str0

Rule argument

pos: right 2

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