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Proof step 38 from proof int2bytes-fillsgn-rec (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 39:
i < ~128 ∨ 128 ≤ i ⊦ fillsgn(# int2bytes(i >> 8) + 1, int2bytes(i >> 8) + i2b(i) ') = int2bytes(i >> 8) + i2b(i) '

Rule argument

Additional information

Used simplifier rules: ⊦ m ≤ m;% ⊦ ¬ m < n ↔ n ≤ m;% ⊦ m < n + 1 ↔ ¬ n < m;% ⊦ m + n ≤ m + n0 ↔ n ≤ n0;% ⊦ m < n ∧ n = n0 → m < n0; (forward)% ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward)% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m < n → (n ≤ m + 1 ↔ n = m + 1);%n0 ≤ n ⊦ n - n0 ≤ m ↔ n ≤ m + n0;% ⊦ ¬ m ≤ n ↔ n < m;% ⊦ m < 1 ↔ m = 0;% ⊦ m < n → n ≠ 0;% ⊦ n ≥ m ↔ m ≤ n;