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Proof step 9 from proof int2bytes-first-neg (previous step)

int2bytes(i) = by ' + bytes, i < 0, 0 ≤ b2i(by), by ∈ int2bytes(i), ~128 ≤ i, ∀ bytes, by, i0. i→n(i0) < i→n(i) → int2bytes(i0) = by ' + bytes ∧ by ∈ int2bytes(i0) ∧ 0 ≤ b2i(by) ∧ i0 < 0 → false ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 10:
~128 ≤ i, i < 128 ⊦ int2bytes((* i)) = i2b(i) '
Proof step 11:
~128 ≤ i ∧ i < 128, i2b(i) ' = by ' + bytes, i < 0, 0 ≤ b2i(by), by ∈ i2b(i) ', ~128 ≤ i, ∀ bytes, by, i0. i→n(i0) < i→n(i) → int2bytes(i0) = by ' + bytes ∧ by ∈ int2bytes(i0) ∧ 0 ≤ b2i(by) ∧ i0 < 0 → false ⊦

Rule argument

name: int2bytes-byte rotate: F subst: [i] → [i]

Additional information

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