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Proof step 6 from proof bytes2int-size-07 (previous step)

1 < #(i2b(~1) ' + by ' + bytes), 0 ≤ (bytes2int((i2b(~1) ' + by ' + bytes).butlast) * 256 + band(b2i((i2b(~1) ' + by ' + bytes).last), 255)) + 2 ^ (8 * (n→i(# bytes) + 1) - 1), 0 ≤ b2i(by), ∀ by, bytes0. # bytes0 < # bytes → 0 ≤ b2i(by) ∧ 0 ≤ bytes2int(i2b(~1) ' + by ' + bytes0) + 2 ^ (8 * (n→i(# bytes0) + 1) - 1) → false ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
0 ≤ b2i(by), 0 ≤ 2 ^ (8 * (n→i(# bytes) + 1) - 1) + band(255, b2i((by ' + bytes).last)) + 256 * bytes2int(i2b(~1) ' + (by ' + bytes).butlast), ∀ by, bytes0. # bytes0 < # bytes → 0 ≤ b2i(by) ∧ 0 ≤ bytes2int(i2b(~1) ' + by ' + bytes0) + 2 ^ (8 * (n→i(# bytes0) + 1) - 1) → false ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ (bytes + by ' + bytes0).butlast = bytes + (by ' + bytes0).butlast;% ⊦ (by ' + by0 ' + bytes).last = (by0 ' + bytes).last;% ⊦ # bytes = 0 ↔ bytes = [];% ⊦ m + n = 1 ↔ m = 0 ∧ n = 1 ∨ n = 0 ∧ m = 1;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;%n < n0 ⊦ (* n) ≠ m0 + (* n0);% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1);% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ # by ' = 0 +1;% ⊦ #(bytes + bytes0) = # bytes + # bytes0;