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Proof step 3 from proof app (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
str2 + str0 = str2 + str1, ∀ str0, str1. str2 + str0 = str2 + str1 → str0 = str1 ⊦ str0 = str1

Rule argument

Additional information

Used simplifier rules: append-rec: ⊦ (char + str) + str0 = char + (str + str0);%inj: ⊦ char + str = char0 + str0 ↔ char = char0 ∧ str = str0; from specification string-data