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Proof step 13 from proof shiftright-neg-mone (previous step)

¬ i < ~1, i→n(j) < # int2bits(i), i < 0, 0 ≤ 2 ^ (abs(j) - 1) + i ⊦ bits2int(firstn(# int2bits(i), mklist(true, i→n(j)) + int2bits(i))) = ~1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
i→n(j) < # int2bits(~1) ⊦ j = 0, bits2int(firstn(# int2bits(~1), mklist(true, i→n(j)) + int2bits(~1))) = ~1

Rule argument

Additional information

Used simplifier rules: ⊦ int2bits(j) ≠ [];% ⊦ # bits = 0 ↔ bits = [];% ⊦ m < n → n ≠ 0; (forward)% ⊦ 0 < abs(i) ↔ i ≠ 0;% ⊦ 0 < 2 ^ i ↔ 0 ≤ i;% ⊦ 1 ≤ i ↔ 0 < i;% ⊦ i ≤ j - i0 ↔ i + i0 ≤ j;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;%i < 0 ⊦ (* i) + j = j - ~ i;% ⊦ i < 0 → (~1 ≤ i ↔ i = ~1);% ⊦ ¬ i < j ↔ j ≤ i;