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Proof step 9 from proof int2bytes-len-eq-pos (previous step)

int2bytes(i) = i2b(0) ' + bytes, 0 ≤ i, 0 ≤ b2i(bytes.first), i2b(0) ∈ int2bytes(i) ⊦ # int2bytes(i) = n ↔ ¬ i < 2 ^ ((n→i(n) - 1) * 8 - 1) ∧ ¬ 2 ^ ((n→i(n) - 1) * 8) ≤ i ∧ n ≠ 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] → [bytes, by, bytes0] rotate: T spec: list-data inst: bytelist name: elim-1

Additional information

Used simplifier rules: int2bytes-zero: ⊦ int2bytes(i) = i2b(0) ' ↔ i = 0;%int2bytes-byte: ~128 ≤ i, i < 128 ⊦ int2bytes((* i)) = i2b(i) ';%no-leading-zero-01: ⊦ 0 ≤ b2i(by) → int2bytes(i) ≠ i2b(0) ' + by ' + bytes;% ⊦ by + bytes = by ' + bytes;% ⊦ bytes + [] = bytes;% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ # by ' = 0 +1;% ⊦ 0 < 2 ^ i ↔ 0 ≤ i;% ⊦ i ≤ j - i0 ↔ i + i0 ≤ j;% ⊦ 1 ≤ i ↔ 0 < i;% ⊦ 0 < i → (0 < i * j ↔ 0 < j);% ⊦ i < j - i0 ↔ i + i0 < j;%0 ≤ i ⊦ (* i) < n→i(n) ↔ i→n(i) < n;% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1);% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i * j = j * i;%⊦ bytes ≠ [] → (by = bytes.first ∧ bytes0 = bytes.rest ↔ bytes = by + bytes0); (elim) Used terms: bytes, by, bytes0