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Proof step 41 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 = [] ⊦ i2b(bits2int(lastn(8, byte2bits(by)))) = by

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 42:
⊦ n = # bits → lastn(n, bits) = bits

Rule argument

seq: ⊦ n = # bits → lastn(n, bits) = bits subst: [bits, n] → [byte2bits(by), 8] rotate: F spec: list-del inst: bitlista name: lastN-all

Additional information

Used simplifier rules: byte2bits-length: ⊦ # byte2bits(by) = 8;%byte2bits-length: ⊦ # byte2bits(by) = 8;%byte2bits-value: ⊦ bits2int(byte2bits(by)) = b2i(by);% ⊦ i2b(b2i(by)) = by; Used terms: byte2bits(by), 8