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Proof step 7 from proof int2bits-len-mon (previous step)

0 ≤ i, 0 ≤ i, i < abs(j), # int2bits(j) < # uint2bits(i) ⊦ j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
# int2bits(j) < # uint2bits(i), i < abs(j), 0 ≤ i ⊦ 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)