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Proof step 2 from proof inj-03 (previous step)

str.char1 = str0.char1, str.rstring = str0.rstring ⊦ str0 = "", str = "", str = str0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 3:
⊦ str ≠ "" → (char = str.char1 ∧ str0 = str.rstring ↔ str = char + str0)
Proof step 4:
str = char + str1, char = str0.char1, str1 = str0.rstring ⊦ str0 = "", str = "", str = str0

Rule argument

spec: string-data inst: name: elim seq: ⊦ str ≠ "" → (char = str.char1 ∧ str0 = str.rstring ↔ str = char + str0) subst: [str, char0, str0] → [str, char0, str1] flags: rewrite, current

Additional information

Used terms: str, char0, str1