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Proof step 58 from proof intshiftrsign-rec-pos (previous step)

¬(mklist(false, i→n(i0)) + int2bits(i / 2) ≠ [] ∧ m ≠ 0), # int2bits(i / 2) = m + 1, 0 < i, 1 < i, 0 ≤ i0, 0 ≤ i, i0 < 31, (sublist(0, m, mklist(false, i→n(i0)) + int2bits(i / 2)) + (mklist(false, i→n(i0)) + int2bits(i / 2))[m] ').first ⊦ bits2uint(sublist(0, m, mklist(false, i→n(i0)) + int2bits(i / 2)) + (mklist(false, i→n(i0)) + int2bits(i / 2))[m] ') = bits2int(sublist(0, m, mklist(false, i→n(i0)) + int2bits(i / 2)) + (mklist(false, i→n(i0)) + int2bits(i / 2))[m] ')

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ bits2int(true ') = ~1;% ⊦ bits2uint(true ') = 1;% ⊦ boolvar '.first ↔ boolvar;% ⊦ [] + bits = bits;% ⊦ sublist(0, 0, bits) = [];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ int2bits(j) ≠ [];% ⊦ i→n(i) = 0 ↔ i = 0;% ⊦ mklist(boolvar, m) = [] ↔ m = 0;% ⊦ ¬ bits + bits0 ≠ [] ↔ bits = [] ∧ bits0 = [];