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Proof step 62 from proof bytes2bytes-zero (previous step)

b2i(by) = 0, b2i(by0) < 0, ∀ by0, by, bytes1. # bytes1 < #(bytes0 + by1 ') → b2i(by) = 0 ∧ b2i(by0) < 0 → bits2bytes(sgn(byte2bits(by) + byte2bits(by0) + bytes2bits(bytes1))) = by ' + by0 ' + bytes1 ⊦ bits2bytes(sgn(false ' + fillsgn(8, int2bits(b2i(by0))) + bytes2bits(bytes0) + byte2bits(by1))) = by ' + by0 ' + bytes0 + by1 ', true ∈ fillsgn(8, int2bits(b2i(by0))), true ∈ bytes2bits(bytes0)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 63:
⊦ bits ≠ [] → (boolvar ∈ fillsgn(n, bits) ↔ boolvar ∈ bits)

Rule argument

seq: ⊦ bits ≠ [] → (boolvar ∈ fillsgn(n, bits) ↔ boolvar ∈ bits) subst: [bits, n, boolvar] → [int2bits(b2i(by0)), 8, true] rotate: F spec: bitopsaux inst: name: fill-in

Additional information

Used simplifier rules: ⊦ int2bits(j) ≠ [];% ⊦ true ∈ int2bits(j) ↔ j ≠ 0;% ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) Used terms: int2bits(b2i(by0)), 8, true