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Proof step 50 from proof bytes2bytes-fix (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 51:
b2i(by0) < 0, # int2bytes(bytes2int(by0 ' + bytes)) < 2 + # bytes, ∀ bytes0. # bytes0 < #(i2b(~1) + 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) ≠ [];%bytes2bytes-first-neg: ⊦ b2i(int2bytes(bytes2int(by ' + bytes)).first) < 0 ↔ b2i(by) < 0;%n < n0 ⊦ (* n) ≠ m0 + (* n0);% ⊦ m < n → n ≠ 0; (forward)% ⊦ i < 0 ∧ 0 ≤ j → i < j;% ⊦ (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;