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Proof step 6 from proof val-trans (previous step)

char0 ' + str < str3, char ' + str0 < char0 ' + str, ∀ str3, char, str0. str < str3 ∧ char ' + str0 < str ∧ (∀ str2, str3. str0 < str2 ∧ str2 < str3 → str0 < str3) → char ' + str0 < str3, ∀ str2, str3. str0 < str2 ∧ str2 < str3 → str0 < str3 ⊦ char ' + str0 < str3

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ char1 ' + str1 < char2 ' + str2 ↔ char1 < char2 ∨ char1 = char2 ∧ str1 < str2
Proof step 8:
char0 ' + str < str3, char < char0 ∨ char = char0 ∧ str0 < str, ∀ str3, char, str0. str < str3 ∧ char ' + str0 < str ∧ (∀ str2, str3. str0 < str2 ∧ str2 < str3 → str0 < str3) → char ' + str0 < str3, ∀ str2, str3. str0 < str2 ∧ str2 < str3 → str0 < str3 ⊦ char ' + str0 < str3

Rule argument

spec: string-less inst: name: Less-rec seq: ⊦ char1 ' + str1 < char2 ' + str2 ↔ char1 < char2 ∨ char1 = char2 ∧ str1 < str2 subst: [str2, char2, str1, char1] → [str, char0, str0, char] flags: rewrite, current

Additional information

Used terms: str, char0, str0, char