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Proof step 6 from proof int2int-lastn-pospos (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: int2bits-not-empty: ⊦ int2bits(j) ≠ [];%int2int-inj: ⊦ bits2int(int2bits(i)) = i;%int2bits-first-02: ⊦ int2bits(j).first ↔ j < 0;% ⊦ i < j → ¬ j < i;% ⊦ m ≤ m;% ⊦ # bits = 0 ↔ bits = [];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ [] + bits = bits;% ⊦ m ≤ n → (n - m = 0 ↔ n = m);% ⊦ # [] = 0;