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Proof step 22 from proof bytes2bytes-mone (previous step)

b2i(by0) = ~1, 0 ≤ b2i(by), ∀ bytes1. # bytes1 < #((by0 + []) + by ') → b2i(bytes1.first) = ~1 ∧ 0 ≤ b2i(bytes1.rest.first) → bits2bytes(sgn(bytes2bits(bytes1))) = bytes1 ∨ # bytes1 = 1 ∨ bytes1 = [] ⊦ false ∈ byte2bits(by0), bits2bytes(sgn(byte2bits(by0) + byte2bits(by))) = by0 ' + by '

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 23:
⊦ byte2bits(by) = fillsgn(8, int2bits(b2i(by)))
Proof step 24:
b2i(by0) = ~1, 0 ≤ b2i(by), ∀ bytes1. # bytes1 < #((by0 + []) + by ') → b2i(bytes1.first) = ~1 ∧ 0 ≤ b2i(bytes1.rest.first) → bits2bytes(sgn(bytes2bits(bytes1))) = bytes1 ∨ # bytes1 = 1 ∨ bytes1 = [] ⊦ false ∈ fillsgn(8, int2bits(b2i(by0))), bits2bytes(sgn(fillsgn(8, int2bits(b2i(by0))) + byte2bits(by))) = by0 ' + by '

Rule argument

name: byte2bits rotate: F subst: [by] → [by0]

Additional information

Used terms: by0