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Proof step 47 from proof int2bits-len-mon (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 48:
j < 0, # uint2bits(~(j + 1)) < # uint2bits(~(i + 1)), i < 0, j < i, j < ~1 ⊦ j = 0

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;