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Proof step 3 from proof append-eqlen-01 (previous step)

∀ 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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
# 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

Rule argument

Additional information

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