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Proof step 26 from proof int2bytes-len-eq-pos (previous step)

int2bytes(i) = i2b(0) ' + bytes, # int2bytes(i) < 2, 0 ≤ i, i < 2 ^ (8 * (1 + n→i(# int2bytes(i) - 1 - 1))), i2b(0) ∈ int2bytes(i) ⊦ bytes = [], # int2bytes(i) = n ↔ ¬ i < 2 ^ ((n→i(n) - 1) * 8 - 1) ∧ ¬ 2 ^ ((n→i(n) - 1) * 8) ≤ i ∧ n ≠ 0

User applied rule cut formula

And obtained the following new goals

Proof step 27:
int2bytes(i) = i2b(0) ' + bytes, # int2bytes(i) < 2, 0 ≤ i, i < 2 ^ (8 * (1 + n→i(# int2bytes(i) - 1 - 1))), i2b(0) ∈ int2bytes(i) ⊦ n = 0, bytes = [], # int2bytes(i) = n ↔ ¬ i < 2 ^ ((n→i(n) - 1) * 8 - 1) ∧ ¬ 2 ^ ((n→i(n) - 1) * 8) ≤ i ∧ n ≠ 0

Rule argument

fma: n = 0

Additional information

Used simplifier rules: int2bytes-empty: ⊦ int2bytes(i) ≠ [];% ⊦ # bytes = 0 ↔ bytes = [];% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j;% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i ≤ j → (¬ ¬ j < i ↔ false); Used formulas: n = 0