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Proof step 3 from proof append-eqlen (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
# str3 + 1 = # str1, (char ' + str3) + str0 = str1 + str2, ∀ str1, str0, str2. # str3 = # str1 ∧ str3 + str0 = str1 + str2 → str3 = str1 ∧ str0 = str2 ⊦ char ' + str3 = str1 ∧ str0 = str2

Rule argument

Additional information

Used simplifier rules: Length-rec: ⊦ #(char ' + str) = # str + 1;%assoc: ⊦ (str1 + str2) + str3 = str1 + str2 + str3; from specification string-append%chartostring-01: ⊦ char + str = char ' + str; from specification string-append