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Proof step 16 from proof intshiftrsign-rec-neg (previous step)

0 ≤ j - 1 ∧ ¬ false ∈ int2bits(31).rest ∧ 0 ≤ 31, 0 ≤ j, 0 < j, i < 0, j < 32 ⊦ bits2int(sublist(0, # int2bits(i), mklist(true, i→n(j)) + int2bits(i))) = bits2int(sublist(0, # int2bits((i - 1) / 2), mklist(true, i→n((j - 1) % (31 + 1))) + int2bits((i - 1) / 2)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ boolvar ∈ boolvar0 ' ↔ boolvar ↔ boolvar0;%⊦ ¬ boolvar ∈ boolvar0 ' + bits ↔ ¬(boolvar ↔ boolvar0) ∧ ¬ boolvar ∈ bits;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ (boolvar ' + bits).rest = bits;% ⊦ bits + boolvar = bits + boolvar ';% ⊦ boolvar + boolvar0 = boolvar ' + boolvar0 ';% ⊦ uint2bits(1) = false + true;%1 < i ⊦ uint2bits((* i)) = uint2bits(i / 2) + (i % 2 = 1);%0 ≤ i ⊦ int2bits((* i)) = uint2bits(i);% ⊦ 1 ≤ i ↔ 0 < i;% ⊦ i ≤ j - i0 ↔ i + i0 ≤ j;