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Proof step 7 from proof int2int-lastn-2pot (previous step)

0 ≤ n→i(m) ∧ j ≠ 0, m ≤ #(int2bits(j) + mklist(false, i→n(n→i(m)))) ⊦ m = 0, bits2int(lastn(m, int2bits(j) + mklist(false, i→n(n→i(m))))) = 0

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: bits2int-zeros-03: ⊦ m ≠ 0 → bits2int(mklist(false, m)) = 0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ lastn(n, bits + mklist(boolvar, n)) = mklist(boolvar, n);% ⊦ i→n(n→i(n)) = n;