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Proof step 16 from proof int2bytes-len-le-neg-03 (previous step)

int2bytes(i).first = i2b(~1), int2bytes(i).rest = [], 0 ≤ b2i([].first), 2 ^ (8 * (n→i(# [].rest) + 1) - 1) + i < 0, 0 ≤ i + 2 ^ ((n→i(# [].rest) + 1) * 8) ⊦ # int2bytes(i) ≤ n ↔ ¬ 2 ^ (8 * n→i(n) - 1) + i < 0

User applied rule insert elim lemma

And closed the goal!

Rule argument

seq: ⊦ bytes ≠ [] → (by = bytes.first ∧ bytes0 = bytes.rest ↔ bytes = by + bytes0) subst: [bytes, by, bytes0] → [int2bytes(i), by, bytes] rotate: T spec: list-data inst: bytelist name: elim-1

Additional information

Used simplifier rules: int2bytes-empty: ⊦ int2bytes(i) ≠ [];%int2bytes-minus-one: ⊦ int2bytes(i) = i2b(~1) ' ↔ i = ~1;%int2bytes-empty: ⊦ int2bytes(i) ≠ [];%int2bytes-byte: ~128 ≤ i, i < 128 ⊦ int2bytes((* i)) = i2b(i) ';% ⊦ by + bytes = by ' + bytes;% ⊦ bytes + [] = bytes;% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ # by ' = 0 +1;%i < 0 ⊦ (* i) + j = j - ~ i;% ⊦ i - j < i0 ↔ i < j + i0;% ⊦ 2 ^ i < 1 ↔ i < 0;% ⊦ 0 < i ∧ 0 ≤ j → ((* i) * i0 < (* j) ↔ i0 < (i + j - 1) / i);% ⊦ n→i(n) < 1 ↔ n = 0;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i * j = j * i;% ⊦ ¬ m ≤ n ↔ n < m;% ⊦ m < 1 ↔ m = 0;%⊦ bytes ≠ [] → (by = bytes.first ∧ bytes0 = bytes.rest ↔ bytes = by + bytes0); (elim) Used terms: int2bytes(i), by, bytes