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Proof step 25 from proof bytes2bytes-first-neg (previous step)

¬ i2b(b2i(by)) ≠ i2b(~1), b2i(by) = ~1, 0 ≤ b2i(int2bytes(bytes2int(by ' + bytes)).first), ∀ by, bytes0. # bytes0 < # bytes → true → (b2i(int2bytes(bytes2int(by ' + bytes0)).first) < 0 ↔ ¬ 0 ≤ b2i(by)) ⊦ bytes = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 26:
0 ≤ b2i(int2bytes(bytes2int(i2b(~1) ' + bytes)).first), ∀ by, bytes0. # bytes0 < # bytes → true → (b2i(int2bytes(bytes2int(by ' + bytes0)).first) < 0 ↔ ¬ 0 ≤ b2i(by)) ⊦ bytes = []

Rule argument

Additional information

Used simplifier rules: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)%~128 ≤ i, i ≤ 127 ⊦ b2i(i2b((* i))) = i;% ⊦ i2b(b2i(by)) = by;