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Proof step 3 from proof uint2bits-len (previous step)

j < 2 ^ (i + 1), 2 ^ i ≤ j, 0 ≤ i, ∀ j0, i. i→n(j0) < i→n(j) → 0 ≤ i ∧ 2 ^ i ≤ j0 ∧ j0 < 2 ^ (i + 1) → # uint2bits(j0) = i→n(i + 2) ⊦ # uint2bits(j) = i→n(i + 2)

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

name: uint2bits-rec-01 rotate: F subst: [i] → [j]

Additional information

Used simplifier rules: uint2bits-one: ⊦ uint2bits(1) = false + true;% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ j ≤ j0 ∧ j0 ≤ j1 → j ≤ j1; (forward)% ⊦ 0 ≤ i → ¬ 2 ^ i ≤ 0;% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i < j ∧ j = i0 → i < i0; (forward)% ⊦ i < j ∧ j < i0 → i +1 < i0; (forward)% ⊦ i < j ∧ j < j0 → i < j0; (forward)% ⊦ boolvar + boolvar0 = boolvar ' + boolvar0 ';% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ # boolvar ' = 0 +1;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ i < (* j) ∧ j < 0 → i < 0; (forward)% ⊦ (* i) < j ∧ 0 < i → 0 ≤ j ∧ 0 < j; (forward)% ⊦ i < j ∧ i = i0 → i0 < j; (forward)% ⊦ i ≤ j ∧ j < j0 → i < j0; (forward)% ⊦ 0 ≤ i → (i < 1 ↔ i = 0);% ⊦ 2 ^ i ≤ 1 ↔ i < 1;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0;%0 ≤ i ⊦ i→n(i) = (* m) ↔ i = n→i(m);% ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0;% ⊦ i ≠ j → (i ≤ j ↔ i < j);% ⊦ i < 1 ↔ i ≤ 0;% ⊦ i < j → ¬ j < i; Used terms: j