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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:
⊦ char1 ' + str1 = char2 ' + str2 ↔ char1 = char2 ∧ str1 = str2
Proof step 8:
∀ 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 ∧ str0 = str, char < char0, char0 < char, char = char0 ∧ str0 < str, char0 = char ∧ str < str0

Rule argument

spec: string-append inst: name: one seq: ⊦ char1 ' + str1 = char2 ' + str2 ↔ char1 = char2 ∧ str1 = str2 subst: [str2, char2, str1, char1] → [str, char0, str0, char] flags: rewrite

Additional information

Used terms: str, char0, str0, char