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Proof step 28 from proof bytes2bytes-first-neg (previous step)

¬(1 < #(i2b(~1) ' + bytes) ∧ b2i((i2b(~1) ' + bytes).first) = ~1 ∧ 0 ≤ b2i((i2b(~1) ' + bytes).rest.first)), 0 ≤ b2i(int2bytes(bytes2int(i2b(~1) ' + 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 29:
b2i(bytes.first) < 0, 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 < 0 ∧ 0 ≤ j → i < j;% ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ (by ' + bytes).rest = bytes;%~128 ≤ i, i ≤ 127 ⊦ b2i(i2b((* i))) = i;% ⊦ (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;