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Proof step 76 from proof bytes2bytes-mone (previous step)

¬ bytes0.rest ≠ [], b2i(bytes0.first) = ~1, 8 = # fillsgn(8, int2bits(b2i(by))), b2i(bytes0.rest.first) < 0, 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 = [] ⊦ bits2bytes(sgn(bytes2bits(bytes0))) = bytes0, bytes0 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 77:
bytes0.rest = [], b2i(bytes0.first) = ~1, 8 = # fillsgn(8, int2bits(b2i(by))), 0 ≤ b2i(by), true ∈ bytes2bits(bytes0), false ∈ bytes2bits(bytes0), b2i([].first) < 0, ∀ bytes1. # bytes1 < #(bytes0 + by ') → b2i(bytes1.first) = ~1 ∧ 0 ≤ b2i(bytes1.rest.first) → bits2bytes(sgn(bytes2bits(bytes1))) = bytes1 ∨ # bytes1 = 1 ∨ bytes1 = [] ⊦ bits2bytes(sgn(bytes2bits(bytes0))) = bytes0, bytes0 = []

Rule argument

Additional information

Used simplifier rules: ⊦ i < 0 ∧ 0 ≤ j → i < j;% ⊦ by '.first = by;% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ [] + bytes = bytes;% ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)