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Proof step 2 from proof append-eqlen-01 (previous step)

# str0 = # str2, str + str0 = str1 + str2 ⊦ str = str1 ∧ str0 = str2

User applied rule structural induction

And obtained the following new goals

Proof step 3:
∀ str0, str2, str1. # str0 = # str2 ∧ str3 + str0 = str1 + str2 → str3 = str1 ∧ str0 = str2, # str0 = # str2, (char + str3) + str0 = str1 + str2 ⊦ char + str3 = str1 ∧ str0 = str2

Rule argument

indvar: str

Additional information

Used simplifier rules: len-app: ⊦ #(str + str0) = # str + # str0;%len-zero: ⊦ # str = 0 ↔ str = "";%string freely generated by "", +;;%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%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2