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Proof step 3 from proof byte-succ (previous step)

b2i(by) < 127, 0 ≤ i, 0 ≤ b2i(by), i < 127, b2i(by) ≤ i ⊦ b2i(i2b(b2i(by) + 1)) = b2i(by) + 1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ 0 ≤ i ∧ i ≤ 127 → b2i(i2b(i)) = i

Rule argument

name: byte-in rotate: F subst: [i] → [b2i(by) + 1]

Additional information

Used simplifier rules: byte-127: ⊦ 126 < b2i(by) ↔ b2i(by) = 127;% ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0;%i0 ≠ 0, i0 ≤ i ⊦ (* i) ≤ j → (* i0) ≤ j;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ i + (* j) ≤ (* j0) ↔ i ≤ j0 - j;% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i ≤ j ∧ j < j0 → i < j0; (forward)% ⊦ i ≤ j ∧ 0 < i → 0 < j; (forward)% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) Used terms: b2i(by) + 1