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Proof step 31 from proof intshiftleft-is-mult-01 (previous step)

¬ ¬ 31 + (j + 1) % 32 < 0, i < 0, j < 0 ⊦ ~31 = (j + 1) % 32, 2 ^ abs(31 + (j + 1) % 32) * (i + 1) = 2 ^ (31 + (j + 1) % 32) * i + 2 ^ abs(31 + (j + 1) % 32), false ∈ int2bits(31).rest

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 32:
i < 0, j < 0, (j + 1) % 32 < ~31, (j + 1) % 32 < 0 ⊦ ~31 = (j + 1) % 32, 0 = i, false ∈ int2bits(31).rest

Rule argument

Additional information

Used simplifier rules: ⊦ i < (* j) ∧ j < 0 → i < 0; (forward)% ⊦ i + (* j) < (* j0) ↔ i < j0 - j;% ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0;% ⊦ j < 0 → (i ^ ~ j = 0 ↔ i = 0);% ⊦ i * j = i ↔ i = 0 ∨ j = 1;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + 0 = i;% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i * 0 = 0;% ⊦ i < 0 → 2 ^ i = 0;% ⊦ i < 0 → abs(i) = ~ i;% ⊦ i + j = j + i;