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Proof step 3 from proof pos-str-01 (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: string-data inst: name: elim seq: ⊦ str ≠ "" → (char = str.char1 ∧ str0 = str.rstring ↔ str = char + str0) subst: [str, char0, str0] → [str, char0, str0] flags:

Additional information

Used simplifier rules: Pos-rec: ⊦ char1 ≠ char2 → pos(char1, char2 ' + str) = pos(char1, str) + 1;%chartostring-01: ⊦ char + str = char ' + str; from specification string-append%assoc: ⊦ (str1 + str2) + str3 = str1 + str2 + str3; from specification string-append%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