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Proof step 9 from proof int2int-lastn-negpos (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
int2bits(i) = bits + lastn(m, int2bits(i)), # bits = # int2bits(i) - m, m ≤ # int2bits(i), i < 0 ⊦ m = 0, i % 2 ^ n→i(m) = 0, bits2int(lastn(m, int2bits(i))) = i % 2 ^ n→i(m) + 2 ^ n→i(m), bits = [], bits.first, lastn(m, int2bits(i)).first

Rule argument

Additional information

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