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Proof step 10 from proof band-011x-is-unsigned-fix (previous step)

¬(¬ false ∈ int2bits(2 ^ i - 1).rest ∧ abs(j) ≤ 2 ^ n→i(# int2bits(2 ^ i - 1) - 2) ∧ 0 < 2 ^ i - 1), # int2bits(j) < i→n(i), 0 < i ⊦ band(j, 2 ^ i - 1) = bits2uint(fillsgn(i→n(i), int2bits(j))), i = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
# int2bits(j) < i→n(i), 0 < i, 2 ^ n→i(1 + i→n(i) - 2) < abs(j) ⊦ band(j, 2 ^ i - 1) = bits2uint(fillsgn(i→n(i), int2bits(j))), i = 0

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ i < j ↔ j ≤ i;% ⊦ 1 < 2 ^ i ↔ 0 < i;% ⊦ i < j - i0 ↔ i + i0 < j;% ⊦ # mklist(boolvar, n) = n;% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i < j → ¬ j < i;% ⊦ ¬ false ∈ mklist(true, n);% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ (boolvar ' + bits).rest = bits;%0 ≤ i ⊦ int2bits(2 ^ i - 1) = false ' + mklist(true, i→n(i));% ⊦ i ≠ j → (i ≤ j ↔ i < j);