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Proof step 5 from proof right-trim-app (previous step)

¬((str0 + str1).last ∈ str ∧ str0 + str1 ≠ ""), ∀ char. char ∈ str1 → char ∈ str, ∀ str0, str, str2. # str2 < # str1 → (∀ char. char ∈ str2 → char ∈ str) → string-right-trim(str, str0 + str2) = string-right-trim(str, str0) ⊦ str1 = "", string-right-trim(str, str0 + str1) = string-right-trim(str, str0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
∀ char. char ∈ str1 → char ∈ str, ∀ str0, str, str2. # str2 < # str1 → (∀ char. char ∈ str2 → char ∈ str) → string-right-trim(str, str0 + str2) = string-right-trim(str, str0) ⊦ str1 = "", string-right-trim(str, str0 + str1) = string-right-trim(str, str0), str1.last ∈ str

Rule argument

Additional information

Used simplifier rules: last-app: ⊦ str1 ≠ "" → (str0 + str1).last = str1.last;% ⊦ str1 + str2 = "" ↔ str1 = "" ∧ str2 = "";% ⊦ (str1 + str2) + str3 = str1 + str2 + str3;