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Proof step 8 from proof hexbyte-byte (previous step)

# str = 2 ⊦ str = "", i2b(hex(str.rstring) + 16 * hexchar(str.char1)) = i2b(hexchar(str.char1) * 16 + hexchar(str.rstring.char1))

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 9:
str = char + str0, # str = 2 ⊦ str = "", i2b(hex(str0) + 16 * hexchar(char)) = i2b(hexchar(char) * 16 + hexchar(str0.char1))

Rule argument

spec: string-data inst: name: elim seq: ⊦ str ≠ "" → (char = str.char1 ∧ str0 = str.rstring ↔ str = char + str0) subst: [str, char0, str0] → [str, char0, str0] flags:

Additional information

Used simplifier rules: ⊦ str ≠ "" → (char = str.char1 ∧ str0 = str.rstring ↔ str = char + str0); (elim) Used terms: str, char0, str0