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Proof step 7 from proof int2bytes-fillsgn-rec (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: int2bytes-empty: ⊦ int2bytes(i) ≠ [];% ⊦ m < n → n ≠ 0;% ⊦ m < n → n ≠ 0; (forward)% ⊦ m + 1 ≤ n ↔ m < n;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ # bytes = 0 ↔ bytes = [];% ⊦ m + n = 1 ↔ m = 0 ∧ n = 1 ∨ n = 0 ∧ m = 1;% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ # by ' = 0 +1;% ⊦ #(bytes + bytes0) = # bytes + # bytes0;% ⊦ bytes + by = bytes + by ';