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Proof step 6 from proof append-eqlen (previous step)

# str3 + 1 = # str1, (char ' + str3) + str0 = str1 + str2, # str3 = # str1.rstring ∧ str3 + str0 = str1.rstring + str2 → str3 = str1.rstring ∧ str0 = str2 ⊦ str1 = "", char ' + str3 = str1 ∧ str0 = str2

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] → [str1, char1, str] flags:

Additional information

Used simplifier rules: Length-rec: ⊦ #(char ' + str) = # str + 1;%append-eqlen-01: ⊦ # str0 = # str2 → (str + str0 = str1 + str2 ↔ str = str1 ∧ str0 = str2);%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%one: ⊦ char1 ' + str1 = char2 ' + str2 ↔ char1 = char2 ∧ str1 = str2; from specification string-append%⊦ str ≠ "" → (char = str.char1 ∧ str0 = str.rstring ↔ str = char + str0); (elim) Used terms: str1, char1, str