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Proof step 5 from proof prefix-num (previous step)

⊦ str0.char1 ' + str1 = "", str0 = "", str0.char1 ' + str1 ⊑ str0 ↔ str1 ⊑ str0.rstring

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 6:
str0 = char + str2 ⊦ char ' + str1 = "", str0 = "", char ' + str1 ⊑ str0 ↔ str1 ⊑ str2

Rule argument

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

Additional information

Used simplifier rules: ⊦ str ≠ "" → (char = str.char1 ∧ str0 = str.rstring ↔ str = char + str0); (elim) Used terms: str0, char0, str