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Proof step 6 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 < char0, char0 < char, char = char0 ∧ str0 < str, char0 = char ∧ str < str0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
str ≠ "" ⊦ char ' + str0 = str + str1 ↔ char = str.char1 ∧ str0 = str.rstring + str1
Proof step 8:
char0 ' ≠ "", ∀ 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 = char0 '.char1 ∧ str0 = char0 '.rstring + str, char < char0, char0 < char, char = char0 ∧ str0 < str, char0 = char ∧ str < str0

Rule argument

spec: list inst: string-new name: firsteq seq: str ≠ "" ⊦ char ' + str0 = str + str1 ↔ char = str.char1 ∧ str0 = str.rstring + str1 subst: [str1, str, str0, char] → [str, char0 ', str0, char] flags: rewrite

Additional information

Used simplifier rules: e: ⊦ char ' ≠ ""; from specification list[string-new] Used terms: str, char0 ', str0, char