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Proof step 21 from proof int2bits-add-low (previous step)

i < n→i(# int2bits(~(j * 2 ^ i + j0))), 0 ≤ 2 ^ n→i(# int2bits(~(j * 2 ^ i + j0))) + j * 2 ^ i + j0, j * 2 ^ i + j0 < 0, 0 < i, j < 0, j0 < 0, 0 < 2 ^ (i - 1) + j0 ⊦ sgn(uint2bits(2 ^ n→i(# int2bits(~(j * 2 ^ i + j0))) + j * 2 ^ i + j0).rest) = int2bits(j - 1) + fillsgn(i→n(i), int2bits(j0))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
0 < 2 ^ (i - 1) + j0, j0 < 0, j < 0, 0 < i, i < n→i(# int2bits(~(j * 2 ^ i + j0))), j * 2 ^ i + j0 < 0, 0 ≤ 2 ^ n→i(# int2bits(~(j * 2 ^ i + j0))) + j * 2 ^ i + j0 ⊦ sgn(uint2bits(2 ^ n→i(# int2bits(~(j * 2 ^ i + j0))) + j * 2 ^ i + j0).rest) = int2bits(j - 1) + fillsgn(i→n(i), int2bits(j0)), j * 2 ^ i + j0 = 0

Rule argument

Additional information

Used simplifier rules: int2bits-not-empty: ⊦ int2bits(j) ≠ [];%int2bits-first-02: ⊦ int2bits(j).first ↔ j < 0;%int2bits-eq: ⊦ int2bits(j) = false ' ↔ j = 0;% ⊦ ~ i = 0 ↔ i = 0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + [] = bits;% ⊦ boolvar + bits = boolvar ' + bits;% ⊦ i < j → ¬ j < i;% ⊦ i * j = j * i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ ~ i < 0 ↔ 0 < i;% ⊦ # bits = 1 ↔ bits = bits.first + [];% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1);% ⊦ # bits = 0 ↔ bits = [];% ⊦ 0 < n ↔ n ≠ 0;%0 ≤ i ⊦ (* i) < n→i(n) ↔ i→n(i) < n;% ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward)% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)% ⊦ i + j = j + i;