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Proof step 8 from proof append-eqlen-01 (previous step)

str1 = char0 + str4, # str0 = # str2, (char ' + str3) + str0 = str1 + str2, ∀ str0, str2, str1. # str0 = # str2 ∧ str3 + str0 = str1 + str2 → str3 = str1 ∧ str0 = str2 ⊦ str1 = "", char = char0 ∧ str3 = str4 ∧ str0 = str2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
str3 + str0 = str4 + str2, # str0 = # str2, ∀ str0, str2, str1. # str0 = # str2 ∧ str3 + str0 = str1 + str2 → str3 = str1 ∧ str0 = str2 ⊦ char0 ' + str4 = "", str3 = str4 ∧ str0 = str2

Rule argument

Additional information

Used simplifier rules: one: ⊦ char1 ' + str1 = char2 ' + str2 ↔ char1 = char2 ∧ str1 = str2; from specification string-append%assoc: ⊦ (str1 + str2) + str3 = str1 + str2 + str3; from specification string-append%chartostring-01: ⊦ char + str = char ' + str; from specification string-append