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Proof step 49 from proof intshiftleft-is-mult-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 50:
0 < i, j < 0, (j + 1) % 32 < ~31, (j + 1) % 32 < 0 ⊦ ~31 = (j + 1) % 32, i = 0, false ∈ int2bits(31).rest

Rule argument

Additional information

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