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Proof step 66 from proof getsetShort-theorem (previous step)

j / 256 = i + 1, 0 ≤ b2i(i2b(j)), b2i(i2b(j)) < 256, ~32768 ≤ j, j < 0, i ≤ 127, ~128 ≤ i ⊦ 256 * i + b2i(i2b(j)) = j, j % 256 = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 67:
⊦ i < 0 ∧ ~128 ≤ i % 256 → b2i(i2b(i)) = i % 256
Proof step 68:
¬(j < 0 ∧ ~128 ≤ j % 256), j / 256 = i + 1, 0 ≤ b2i(i2b(j)), b2i(i2b(j)) < 256, ~32768 ≤ j, j < 0, i ≤ 127, ~128 ≤ i ⊦ 256 * i + b2i(i2b(j)) = j, j % 256 = 0

Rule argument

spec: inst: name: bib-neg-high seq: ⊦ i < 0 ∧ ~128 ≤ i % 256 → b2i(i2b(i)) = i % 256 subst: [i] → [j] flags: rewrite, current

Additional information

Used simplifier rules: numleledel-01: i0 ≠ 0, i0 ≤ i ⊦ (* i) ≤ j → (* i0) ≤ j; from specification int-abs%mod-ngez: ⊦ 0 < j ∧ i < 0 → (0 ≤ i % j ↔ i % j = 0); from specification int-div Used terms: j