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Proof step 6 from proof bytes2int-rec-first-zero (previous step)

1 < #(i2b(0) ' + by ' + bytes), ∀ by, bytes0. # bytes0 < # bytes → true → bytes0 = [] ∨ bytes2int(i2b(0) ' + by ' + bytes0) = band(b2i(by), 255) * 256 ^ n→i(# bytes0) + bytes2int(i2b(0) ' + bytes0) ⊦ bytes2int((i2b(0) ' + by ' + bytes).butlast) * 256 + band(b2i((i2b(0) ' + by ' + bytes).last), 255) = band(b2i(by), 255) * 256 ^ n→i(# bytes) + bytes2int(i2b(0) ' + bytes), bytes = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
∀ by, bytes0. # bytes0 < # bytes → true → bytes0 = [] ∨ bytes2int(i2b(0) ' + by ' + bytes0) = band(b2i(by), 255) * 256 ^ n→i(# bytes0) + bytes2int(i2b(0) ' + bytes0) ⊦ bytes = [], band(255, b2i(bytes.last)) + 256 * bytes2int((i2b(0) ' + by ') + bytes.butlast) = band(b2i(by), 255) * 256 ^ n→i(# bytes) + bytes2int(i2b(0) ' + bytes)

Rule argument

Additional information

Used simplifier rules: ⊦ # 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);% ⊦ # by ' = 0 +1;% ⊦ #(bytes + bytes0) = # bytes + # bytes0;% ⊦ bytes ≠ [] → (bytes0 + bytes).butlast = bytes0 + bytes.butlast;% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ bytes ≠ [] → (by ' + bytes).last = bytes.last;% ⊦ (by ' + by0 ' + bytes).last = (by0 ' + bytes).last;