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Proof step 13 from proof int2bytes-fillsgn-rec (previous step)

¬(# int2bytes(i >> 8) < n - 1 ∧ bytes2int(int2bytes(i >> 8)) < 0), # int2bytes(i >> 8) < n, # int2bytes(i >> 8) + 1 < n, i < ~128 ∨ 128 ≤ i ⊦ fillsgn(n, int2bytes(i >> 8) + i2b(i) ') = fillsgn(n - 1, int2bytes(i >> 8)) + i2b(i) ', n = # int2bytes(i >> 8) + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
# int2bytes(i >> 8) < n, # int2bytes(i >> 8) + 1 < n, 128 ≤ i, 0 ≤ i >> 8, 0 < i, 0 ≤ i ⊦ fillsgn(n, int2bytes(i >> 8) + i2b(i) ') = fillsgn(n - 1, int2bytes(i >> 8)) + i2b(i) ', n = # int2bytes(i >> 8) + 1

Rule argument

Additional information

Used simplifier rules: int2int: ⊦ bytes2int(int2bytes(i)) = i;% ⊦ i ≤ j ∧ 0 < i → 0 < j; (forward)%i0 ≠ 0, i0 ≤ i ⊦ (* i) ≤ j → (* i0) ≤ j;% ⊦ i < 0 → ¬ 0 ≤ i >> j;% ⊦ i < (* j) ∧ j < 0 → i < 0; (forward)% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m < n → (n ≤ m ↔ false);% ⊦ ¬ m < n ↔ n ≤ m;% ⊦ m < n + 1 ↔ ¬ n < m;% ⊦ m ≠ n → (¬ m < n ↔ n < m);% ⊦ m + n = n + m;%n0 ≤ n ⊦ m < n - n0 ↔ m + n0 < n;% ⊦ ¬ m ≤ n ↔ n < m;% ⊦ m < 1 ↔ m = 0;% ⊦ m < n → n ≠ 0;