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Proof step 13 from proof bytes2bytes-fix (previous step)

¬(1 < #(by ' + bytes1) ∧ b2i((by ' + bytes1).first) = 0 ∧ b2i((by ' + bytes1).rest.first) < 0), b2i(by) = 0, ∀ bytes0. # bytes0 < #(by + bytes1) → true → bytes0 = [] ∨ fix(# bytes0, int2bytes(bytes2int(bytes0))) = bytes0 ⊦ fix(1 + # bytes1, int2bytes(bytes2int(by ' + bytes1))) = by ' + bytes1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
b2i(by) = 0, 0 ≤ b2i(bytes1.first), ∀ bytes0. # bytes0 < #(by + bytes1) → true → bytes0 = [] ∨ fix(# bytes0, int2bytes(bytes2int(bytes0))) = bytes0 ⊦ fix(1 + # bytes1, int2bytes(bytes2int(by ' + bytes1))) = by ' + bytes1, bytes1 = []

Rule argument

Additional information

Used simplifier rules: bytes2int-one: ⊦ bytes2int(by ') = b2i(by);%int2bytes-byte-01: ⊦ int2bytes(b2i(by)) = by ';%fix-one: ⊦ fix(1, by ') = by ';% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ by ' = by0 ' ↔ by = by0;% ⊦ bytes + [] = bytes;% ⊦ # [] = 0;% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ (by ' + bytes).rest = bytes;% ⊦ (by ' + bytes).first = by;% ⊦ # bytes = 0 ↔ bytes = [];% ⊦ 1 = m + 1 ↔ m = 0;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + 1 ≠ 0;% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1);% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ # by ' = 0 +1;% ⊦ #(bytes + bytes0) = # bytes + # bytes0;