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Proof step 11 from proof int2int-lastn-posneg (previous step)

# lastn(m, int2bits(i)) ≠ 0 ∧ # bits ≠ 0 ∧ ¬ bits.first, bits2int(int2bits(i)) = bits2uint(bits) * 2 ^ n→i(# lastn(m, int2bits(i))) + bits2uint(lastn(m, int2bits(i))), # bits = # int2bits(i) - m, int2bits(i) = bits + lastn(m, int2bits(i)), m ≤ # int2bits(i), lastn(m, int2bits(i)).first, 0 < i ⊦ bits = [], 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 12:
i = bits2uint(lastn(m, int2bits(i))) + 2 ^ n→i(m) * bits2uint(bits), int2bits(i) = bits + lastn(m, int2bits(i)), # bits = # int2bits(i) - m, 0 < i, lastn(m, int2bits(i)).first, m ≤ # int2bits(i) ⊦ m = 0, i % 2 ^ n→i(m) = bits2int(lastn(m, int2bits(i))) + 2 ^ n→i(m), bits = [], bits.first

Rule argument

Additional information

Used simplifier rules: int2int-inj: ⊦ bits2int(int2bits(i)) = i;% ⊦ i < j ∧ j = i0 → i < i0; (forward)% ⊦ # bits = 0 ↔ bits = [];% ⊦ n ≤ # bits → # lastn(n, bits) = n;