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Proof step 3 from proof int2bits-fillsgn (previous step)

∀ i. mklist(¬ 0 ≤ i, m) + int2bits(i) = fillsgn(m + # int2bits(i), int2bits(i)) ⊦ mklist(¬ 0 ≤ i, m +1) + int2bits(i) = fillsgn(m +1 + # int2bits(i), int2bits(i))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
∀ i0. mklist(i0 < 0, m) + int2bits(i0) = fillsgn(m + # int2bits(i0), int2bits(i0)) ⊦ ((¬ 0 ≤ i)' + mklist(¬ 0 ≤ i, m)) + int2bits(i) = fillsgn(# int2bits(i) + m + 1, int2bits(i))

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ i ≤ j ↔ j < i;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ mklist(boolvar, n) + boolvar ' = boolvar ' + mklist(boolvar, n);% ⊦ mklist(boolvar, 1) = boolvar ';% ⊦ mklist(boolvar, m + n) = mklist(boolvar, m) + mklist(boolvar, n);% ⊦ m +1 = m + 1;