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Proof step 2 from proof char-in (previous step)

∀ char. char ∈ l→s(chars0) ↔ char ∈ chars0 ⊦ char ∈ l→s(char0 + chars0) ↔ char ∈ char0 + chars0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ char. char ∈ l→s(chars0) ↔ char ∈ chars0 ⊦ char ∈ char0 ' + l→s(chars0) ↔ char = char0 ∨ char ∈ chars0

Rule argument

Additional information

Used simplifier rules: ⊦ char ∈ char0 ' ↔ char = char0;% ⊦ char ∈ chars + chars0 ↔ ¬ ¬(char ∈ chars ∨ char ∈ chars0);% ⊦ (chars + chars0) + chars1 = chars + chars0 + chars1;% ⊦ char + str = char ' + str;% ⊦ l→s(char ' + chars) = char + l→s(chars);% ⊦ char + chars = char ' + chars;