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Proof step 27 from proof int2bytes-byte (previous step)

0 ≤ i ∧ 1 < n0 + 8, 2 ^ (n→i(n0 + 8) - 2) ≤ i ∧ i < 2 ^ (n→i(n0 + 8) - 1), i < 128, ~128 ≤ i ⊦ n0 = 0, bits2bytes(firstn(n0, int2bits(i))) + i2b(bits2int(lastn(8, int2bits(i)))) ' = i2b(i) '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ls-01: ⊦ i + j < i + i0 ↔ j < i0; from specification int-basic2%ls-pos-2pot-01: 0 ≤ j ⊦ 2 ^ j < 2 ^ j0 ↔ j < j0; from specification int-pot%num-n2i-ls-07: i < 0 ⊦ (* i) ≤ n→i(n); from specification intnat%numle-02: ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0; from specification int-abs%pot2-pos: ⊦ 0 < 2 ^ i ↔ 0 ≤ i; from specification int-pot%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%numleledel-01: i0 ≠ 0, i0 ≤ i ⊦ (* i) ≤ j → (* i0) ≤ j; from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%numsub: ⊦ (* i) + i0 - (* j) = (i - j) + i0; from specification int-abs%n2i-add: ⊦ n→i(m + n) = n→i(m) + n→i(n); from specification intnat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%add-num-03: n < n0 ⊦ (* n) ≠ m0 + (* n0); from specification nat%one: ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); from specification nat-basic1