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Proof step 33 from proof int2bits-len-mon (previous step)

0 ≤ ~(i + 1), i < j, 0 < j + i, i < 0, # int2bits(j) < # uint2bits(~(i + 1)), 0 ≤ j, 1 < j, i + 1 < j, 0 < j ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 34:
0 < j, i + 1 < j, 1 < j, 0 ≤ j, # int2bits(j) < # uint2bits(~(i + 1)), i < 0, 0 < j + i, i < j ⊦

Rule argument

Additional information

Used simplifier rules: uint2bits-empty: 0 ≤ i ⊦ uint2bits(i) ≠ [];% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i < j → (¬ ¬ j ≤ i ↔ false);% ⊦ # bits = 0 ↔ bits = [];% ⊦ m < n → n ≠ 0; (forward)% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i ≤ ~1 ↔ i < 0;% ⊦ i + (* j) ≤ (* j0) ↔ i ≤ j0 - j;% ⊦ i ≤ ~ j ↔ i + j ≤ 0;