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Proof step 15 from proof int2bytes-byte (previous step)

# int2bits(i) = n0 + 8, ~128 ≤ i, i < 0 ⊦ n0 = 0, bits2bytes(firstn(n0, int2bits(i))) + i2b(bits2int(lastn(8, int2bits(i)))) ' = i2b(i) '

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 16:
⊦ j < 0 ∧ 1 < n → (# int2bits(j) = n ↔ ~(2 ^ (n→i(n) - 1)) ≤ j ∧ j < ~(2 ^ (n→i(n) - 2)))
Proof step 17:
i < 0 ∧ 1 < n0 + 8, ~(2 ^ (n→i(n0 + 8) - 1)) ≤ i ∧ i < ~(2 ^ (n→i(n0 + 8) - 2)), ~128 ≤ i, i < 0 ⊦ n0 = 0, bits2bytes(firstn(n0, int2bits(i))) + i2b(bits2int(lastn(8, int2bits(i)))) ' = i2b(i) '

Rule argument

spec: int2bits inst: name: int2bits-leneq-neg seq: ⊦ j < 0 ∧ 1 < n → ( # int2bits(j) = n ↔ ~(2 ^ (n→i(n) - 1)) ≤ j ∧ j < ~(2 ^ (n→i(n) - 2))) subst: [n, j] → [n0 + 8, i] flags: rewrite

Additional information

Used simplifier rules: zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs%one: ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); from specification nat-basic1%add-num-03: n < n0 ⊦ (* n) ≠ m0 + (* n0); from specification nat%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2 Used terms: n0 + 8, i