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Proof step 11 from proof uint2bits-len-mon (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
0 < j, # uint2bits(j / 2) < # uint2bits(i / 2), i ≤ j, 0 ≤ i, 0 ≤ j, 0 < i, ∀ j0, i. i→n(j0) < i→n(j) → 0 ≤ j0 ∧ # uint2bits(j0) < # uint2bits(i) ∧ i ≤ j0 ∧ 0 ≤ i → false ⊦ i = 1, j = 1

Rule argument

Additional information

Used simplifier rules: uint2bits-empty: 0 ≤ i ⊦ uint2bits(i) ≠ [];% ⊦ 0 ≤ i / 2 ↔ ~1 ≤ i;% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i < j ∧ ¬ i0 < j → ¬ i0 < i;% ⊦ # bits = 0 ↔ bits = [];% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ m < n → n ≠ 0; (forward)% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n < m + n0 ↔ n < n0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ bits + boolvar = bits + boolvar ';