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Proof step 17 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(bits0) * 2 ^ n→i(m) + bits2uint(lastn(m, int2bits(i)))) % 2 ^ n→i(m) = bits2int(lastn(m, int2bits(i))) + 2 ^ n→i(m), m = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
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, 2 ^ n→i(m) ≤ bits2uint(lastn(m, int2bits(i))), 0 ≤ i ⊦ i % 2 ^ n→i(m) = bits2int(lastn(m, int2bits(i))) + 2 ^ n→i(m), m = 0

Rule argument

Additional information

Used simplifier rules: bits2uint-not-neg-01: ⊦ ¬ bits2uint(bits) < 0;%int2bits-first: ⊦ j < 0 → int2bits(j) ≠ false ' + bits;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ i < j ∧ j < j0 → i < j0; (forward)% ⊦ i < j ∧ j < i0 → i +1 < i0; (forward)% ⊦ i < j ∧ i = i0 → i0 < j; (forward)% ⊦ i + j = j + i;% ⊦ i * j = j * i;% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ ¬ i ≤ j ↔ j < i;