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Proof step 47 from proof shiftzero-pos (previous step)

# int2bits(i) ≠ i→n(j), # int2bits(i) < n, 0 ≤ i, i→n(j) ≤ # int2bits(i), i→n(j) < n ⊦ bits2int(firstn(n, mklist(false, i→n(j)) + mklist(false, n - # int2bits(i)) + int2bits(i))) = bits2int(mklist(false, i→n(j)) + firstn(# int2bits(i) - i→n(j), int2bits(i))), # int2bits(i) ≠ i→n(j) ∧ ¬ firstn(# int2bits(i) - i→n(j), int2bits(i)).first

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 48:
firstn(# int2bits(i) - i→n(j), int2bits(i)).first, i→n(j) < # int2bits(i), 0 ≤ i, # int2bits(i) < n, i→n(j) < n, i→n(j) + 1 < n ⊦ # int2bits(i) = i→n(j), bits2int(firstn(n, mklist(false, i→n(j)) + mklist(false, n - # int2bits(i)) + int2bits(i))) = bits2int(mklist(false, i→n(j)) + firstn(# int2bits(i) - i→n(j), int2bits(i)))

Rule argument

Additional information

Used simplifier rules: ⊦ m +1 = m + 1;% ⊦ int2bits(j) ≠ [];% ⊦ # bits = 0 ↔ bits = [];% ⊦ m < n → n ≠ 0; (forward)% ⊦ n < n0 ∧ n0 < n1 → n +1 < n1; (forward)% ⊦ m ≠ n → (m ≤ n ↔ m < n);