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Proof step 6 from proof int2bytes-len-eq-pos (previous step)

int2bytes(i).first = i2b(0), 0 ≤ i ⊦ ¬ 0 ≤ b2i(int2bytes(i).rest.first), # int2bytes(i) = n ↔ ¬ i < 2 ^ ((n→i(n) - 1) * 8 - 1) ∧ ¬ 2 ^ ((n→i(n) - 1) * 8) ≤ i ∧ n ≠ 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
int2bytes(i).first = i2b(0), 0 ≤ i, 0 ≤ b2i(int2bytes(i).rest.first) ⊦ # int2bytes(i) = n ↔ ¬ i < 2 ^ ((n→i(n) - 1) * 8 - 1) ∧ ¬ 2 ^ ((n→i(n) - 1) * 8) ≤ i ∧ n ≠ 0

Rule argument

Additional information

Used simplifier rules: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)