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Proof step 4 from proof byte-mult (previous step)

⊦ b2i(i2b(i * 256)) = b2i(i2b(0))

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ ~128 ≤ i ∧ i ≤ 127 → b2i(i2b(i)) = i
Proof step 6:
¬(~128 ≤ i * 256 ∧ i * 256 ≤ 127) ⊦ b2i(i2b(i * 256)) = b2i(i2b(0))

Rule argument

name: bib-in-01 rotate: F subst: [i] → [i * 256]

Additional information

Used simplifier rules: byte: ⊦ b2i(i2b(0)) = 0;% ⊦ i * j = 0 ↔ i = 0 ∨ j = 0;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i * j = j * i;% ⊦ 0 < i ∧ j < 0 → ((* j) ≤ (* i) * i0 ↔ j / i ≤ i0);% ⊦ i ≠ j → (i ≤ j ↔ i < j);% ⊦ 0 < i ∧ 0 ≤ j → ((* i) * i0 ≤ (* j) ↔ i0 ≤ j / i);% ⊦ i < j → ¬ j < i; Used terms: i * 256