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Proof step 13 from proof int2bytes-index-rightshift-02 (previous step)

0 ≤ i1, i1 ≤ n→i(# int2bytes(i)), i1 ≤ 4, 0 < i1, # int2bytes(i) ≤ # int2bytes(i) - i→n(i1) ⊦ int2bytes(i)[# int2bytes(i) - i→n(i1)] = i2b(i >> 8 * (i1 - 1))

User applied rule insert elim lemma

And obtained the following new goals

Rule argument

seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [# int2bytes(i), i→n(i1), n] rotate: T spec: nat inst: name: elim-sub

Additional information

Used simplifier rules: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward)% ⊦ n→i(m + n) = n→i(m) + n→i(n);% ⊦ 0 < i → n→i(i→n(i)) = i;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m;% ⊦ m + m0 ≤ m ↔ m0 = 0;% ⊦ i→n(i) = 0 ↔ i = 0;% ⊦ i ≤ i + j ↔ 0 ≤ j;% ⊦ 0 ≤ n→i(n);% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ m ≤ m + n;% ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)% ⊦ 0 + m = m;% ⊦ i + 0 = i;%0 ≤ i ⊦ (* i) < n→i(n) ↔ i→n(i) < n;% ⊦ 0 < n ↔ n ≠ 0;% ⊦ 0 ≤ m;%n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: # int2bytes(i), i→n(i1), n