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Proof step 4 from proof append-eqlen-01 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ str1 ≠ "" → (char ' + str = str1 ↔ char = str1.char1 ∧ str = str1.rstring)
Proof step 6:
str1 ≠ "", # str0 = # str2, (char ' + str3) + str0 = str1 + str2, ∀ str0, str2, str1. # str0 = # str2 ∧ str3 + str0 = str1 + str2 → str3 = str1 ∧ str0 = str2 ⊦ (char = str1.char1 ∧ str3 = str1.rstring) ∧ str0 = str2

Rule argument

spec: string-append inst: name: rec seq: ⊦ str1 ≠ "" → (char ' + str = str1 ↔ char = str1.char1 ∧ str = str1.rstring) subst: [str1, str, char] → [str1, str3, char] flags: rewrite

Additional information

Used simplifier rules: Length-rec: ⊦ #(char ' + str) = # str + 1;%len-app: ⊦ #(str + str0) = # str + # str0;%len-zero: ⊦ # str = 0 ↔ str = "";%append-base: ⊦ "" + str = str; from specification string-append%assoc: ⊦ (str1 + str2) + str3 = str1 + str2 + str3; from specification string-append%eq: ⊦ m = m + n ↔ n = 0; from specification nat-basic2%z-01: ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2 Used terms: str1, str3, char