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Proof step 29 from proof bytes2bytes-fix (previous step)

0 ≤ b2i(int2bytes(bytes2int(by0 ' + bytes)).first) ∧ # int2bytes(bytes2int(by0 ' + bytes)) < 2 + # bytes ∧ int2bytes(bytes2int(by0 ' + bytes)) ≠ [], 0 ≤ b2i(by0), 0 ≤ b2i(by0), ∀ bytes0. # bytes0 < #(i2b(0) + by0 + bytes) → true → bytes0 = [] ∨ fix(# bytes0, int2bytes(bytes2int(bytes0))) = bytes0 ⊦ i2b(0) ' + fix((2 + # bytes)-1, int2bytes(bytes2int(by0 ' + bytes))) = i2b(0) ' + by0 ' + bytes

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 30:
0 ≤ b2i(int2bytes(bytes2int(by0 ' + bytes)).first), 0 ≤ b2i(by0), # int2bytes(bytes2int(by0 ' + bytes)) < 2 + # bytes, ∀ bytes0. # bytes0 < #(i2b(0) + by0 + bytes) → true → bytes0 = [] ∨ fix(# bytes0, int2bytes(bytes2int(bytes0))) = bytes0 ⊦ fix(1 + # bytes, int2bytes(bytes2int(by0 ' + bytes))) = by0 ' + bytes

Rule argument

Additional information

Used simplifier rules: int2bytes-empty: ⊦ int2bytes(i) ≠ [];%n < n0 ⊦ (* n) ≠ m0 + (* n0);% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ m < n → n ≠ 0; (forward)% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ bytes + bytes0 = bytes + bytes1 ↔ bytes0 = bytes1;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ n ≤ m → m + n0 - n = (m - n) + n0;% ⊦ n -1 = n - 1;