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Proof step 53 from proof bytes2bytes-mone (previous step)

b2i(by0) = ~1, 0 ≤ b2i(by1), ∀ bytes1. # bytes1 < #((by0 + by1 + bytes0) + by ') → b2i(bytes1.first) = ~1 ∧ 0 ≤ b2i(bytes1.rest.first) → bits2bytes(sgn(bytes2bits(bytes1))) = bytes1 ∨ # bytes1 = 1 ∨ bytes1 = [] ⊦ bits2bytes(sgn(true ' + fillsgn(8, int2bits(b2i(by1))) + bytes2bits(bytes0) + byte2bits(by))) = by0 ' + (by1 ' + bytes0) + by ', false ∈ fillsgn(8, int2bits(b2i(by1))), false ∈ bytes2bits(bytes0)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 54:
⊦ bits ≠ [] → (boolvar ∈ fillsgn(n, bits) ↔ boolvar ∈ bits)
Proof step 55:
int2bits(b2i(by1)) ≠ [], b2i(by0) = ~1, 0 ≤ b2i(by1), ∀ bytes1. # bytes1 < #((by0 + by1 + bytes0) + by ') → b2i(bytes1.first) = ~1 ∧ 0 ≤ b2i(bytes1.rest.first) → bits2bytes(sgn(bytes2bits(bytes1))) = bytes1 ∨ # bytes1 = 1 ∨ bytes1 = [] ⊦ bits2bytes(sgn(true ' + fillsgn(8, int2bits(b2i(by1))) + bytes2bits(bytes0) + byte2bits(by))) = by0 ' + (by1 ' + bytes0) + by ', false ∈ int2bits(b2i(by1)), false ∈ bytes2bits(bytes0)

Rule argument

seq: ⊦ bits ≠ [] → (boolvar ∈ fillsgn(n, bits) ↔ boolvar ∈ bits) subst: [bits, n, boolvar] → [int2bits(b2i(by1)), 8, false] rotate: F spec: bitopsaux inst: name: fill-in

Additional information

Used simplifier rules: ⊦ int2bits(j) ≠ []; Used terms: int2bits(b2i(by1)), 8, false