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Proof step 3 from proof pos-str-02 (previous step)

⊦ pos(char, str) = pos(char, str.rstring) + 1, char = str.char1, str = ""

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

spec: list-data inst: string-new name: elim-1 seq: ⊦ str ≠ "" → (char = str.char1 ∧ str0 = str.rstring ↔ str = char + str0) subst: [str, char0, str0] → [str, char0, str0] flags:

Additional information

Used simplifier rules: c: ⊦ char + str = char ' + str; from specification list[string-new]%pos-n: ⊦ char ≠ char0 → pos(char, char0 ' + str) = pos(char, str) + 1; from specification list-del[string-new]%a: ⊦ (str + str0) + str1 = str + str0 + str1; from specification list[string-new]%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%⊦ str ≠ "" → (char = str.char1 ∧ str0 = str.rstring ↔ str = char + str0); (elim) Used terms: str, char0, str0