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Proof step 5 from proof val-total (previous step)

∀ str0, char. (∀ str2. str0 = str2 ∨ str0 < str2 ∨ str2 < str0) → char ' + str0 = str ∨ char ' + str0 < str ∨ str < char ' + str0, ∀ str2. str0 = str2 ∨ str0 < str2 ∨ str2 < str0 ⊦ char ' + str0 = char0 + str, char ' + str0 < char0 + str, char0 + str < char ' + str0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
∀ str0, char. (∀ str2. str0 = str2 ∨ str0 < str2 ∨ str2 < str0) → char ' + str0 = str ∨ char ' + str0 < str ∨ str < char ' + str0, ∀ str2. str0 = str2 ∨ str0 < str2 ∨ str2 < str0 ⊦ char ' + str0 = char0 ' + str, char < char0, char0 < char, char = char0 ∧ str0 < str, char0 = char ∧ str < str0

Rule argument

Additional information

Used simplifier rules: Less-rec: ⊦ char1 ' + str1 < char2 ' + str2 ↔ char1 < char2 ∨ char1 = char2 ∧ str1 < str2;%a: ⊦ (str + str0) + str1 = str + str0 + str1; from specification list[string-new]%c: ⊦ char + str = char ' + str; from specification list[string-new]