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Proof step 17 from proof int2bytes-first-zerobyte (previous step)

i ≤ 127, 0 < i, ∀ i0. i→n(i0) < i→n(i) → 0 ≤ i0 → (int2bytes(i0).first = i2b(0) ↔ n→i(# int2bits(i0)) % 8 = 1) ⊦ i = 0, i2b(i) = i2b(0) ↔ n→i(# int2bits(i)) % 8 = 1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 18:
~128 ≤ i, i ≤ 127, ~128 ≤ j, j ≤ 127 ⊦ i2b((* i)) = i2b((* j)) ↔ i = j
Proof step 19:
~128 ≤ i ∧ i ≤ 127 ∧ ~128 ≤ 0 ∧ 0 ≤ 127, i ≤ 127, 0 < i, ∀ i0. i→n(i0) < i→n(i) → 0 ≤ i0 → (int2bytes(i0).first = i2b(0) ↔ n→i(# int2bits(i0)) % 8 = 1) ⊦ i = 0, i = 0 ↔ n→i(# int2bits(i)) % 8 = 1

Rule argument

seq: ~128 ≤ i, i ≤ 127, ~128 ≤ j, j ≤ 127 ⊦ i2b((* i)) = i2b((* j)) ↔ i = j subst: [j, i] → [0, i] rotate: F spec: byte inst: name: num-i2b-equal

Additional information

Used simplifier rules: ⊦ ¬ i ≤ j ↔ j < i;%j0 ≠ 0, j ≤ j0 ⊦ i ≤ (* j) → i ≤ (* j0);% ⊦ i < j ∧ ¬ i0 < j → ¬ i0 < i; Used terms: 0, i