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Proof step 50 from proof shiftzero-pos (previous step)

# int2bits(i) < n, 0 ≤ i, i→n(j) ≤ # int2bits(i), i→n(j) < 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))), # 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 51:
# int2bits(i) = i→n(j), 0 ≤ i, i→n(j) < n ⊦ bits2int(firstn(n, mklist(false, i→n(j)) + mklist(false, n - i→n(j)) + int2bits(i))) = bits2int(mklist(false, i→n(j)))

Rule argument

Additional information

Used simplifier rules: ⊦ i ≤ i;% ⊦ i→n(i) ≤ i→n(j) ↔ abs(i) ≤ abs(j);% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + [] = bits;% ⊦ firstn(0, bits) = [];% ⊦ m - m = 0;% ⊦ m < n ∧ m = n0 → n0 < n; (forward)