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Proof step 67 from proof bytes2bytes-mone (previous step)

b2i(bytes0.first) = ~1, true ∈ bytes2bits(bytes0), false ∈ bytes2bits(bytes0), 0 ≤ b2i((bytes0.rest + by ').first), ∀ bytes1. # bytes1 < #(bytes0 + by ') → b2i(bytes1.first) = ~1 ∧ 0 ≤ b2i(bytes1.rest.first) → bits2bytes(sgn(bytes2bits(bytes1))) = bytes1 ∨ # bytes1 = 1 ∨ bytes1 = [] ⊦ 8 = # fillsgn(8, int2bits(b2i(by))), bits2bytes(sgn(bytes2bits(bytes0))) + i2b(bits2int(lastn(8, sgn(bytes2bits(bytes0)) + fillsgn(8, int2bits(b2i(by)))))) ' = bytes0 + by ', bytes0 = []

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 68:
⊦ # bits ≤ n → # fillsgn(n, bits) = n

Rule argument

seq: ⊦ # bits ≤ n → # fillsgn(n, bits) = n subst: [bits, n] → [int2bits(b2i(by)), 8] rotate: F spec: bitopsaux inst: name: fill-length

Additional information

Used simplifier rules: ⊦ ¬ m ≤ n ↔ n < m;% ⊦ ¬ 8 < # int2bits(b2i(by)); Used terms: int2bits(b2i(by)), 8