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Proof step 10 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) ') = fillsgn(n - 1, int2bytes(i >> 8)) + i2b(i) '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
# 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

Rule argument

Additional information

Used simplifier rules: ⊦ m < n ∧ m + 1 ≠ n → m + 1 < n; (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;