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Proof step 20 from proof int2int-lastn-posneg (previous step)

0 ≤ bits2uint(bits0) * 2 ^ n→i(m) + bits2uint(lastn(m, int2bits(i))) ∧ 0 ≤ bits2uint(lastn(m, int2bits(i))) ∧ bits2uint(lastn(m, int2bits(i))) < 2 ^ n→i(m), 1 + # bits0 = # int2bits(i) - m, int2bits(i) = (false ' + bits0) + lastn(m, int2bits(i)), i = bits2uint(bits0) * 2 ^ n→i(m) + bits2uint(lastn(m, int2bits(i))), false ∈ int2bits(i), m ≤ # int2bits(i), lastn(m, int2bits(i)).first, 0 < i ⊦ bits2uint(lastn(m, int2bits(i))) = bits2int(lastn(m, int2bits(i))) + 2 ^ n→i(m), m = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward)% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ 0 ≤ j1 ∧ j1 < j → (0 ≤ i0 * j + j1 ↔ 0 ≤ i0);