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Proof step 3 from proof fill-sgn-01 (previous step)

∀ n. # bits0 ≤ n → fillsgn(n, sgn(bits0)) = fillsgn(n, bits0) ⊦ #(boolvar + bits0) ≤ n → fillsgn(n, sgn(boolvar + bits0)) = fillsgn(n, boolvar + bits0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
# bits0 < n, ∀ n0. ¬ n0 < # bits0 → fillsgn(n0, sgn(bits0)) = fillsgn(n0, bits0) ⊦ fillsgn(n, sgn(boolvar ' + bits0)) = mklist(boolvar, n - # bits0) + bits0

Rule argument

Additional information

Used simplifier rules: fillsgn-fill: ⊦ # bits < n → fillsgn(n, boolvar ' + bits) = mklist(boolvar, n - # bits) + bits;% ⊦ m < n → n ≠ 0;% ⊦ m < n → n ≠ 0; (forward)% ⊦ ¬ m ≤ n ↔ n < m;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + 1 ≤ n ↔ m < n;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ boolvar + bits = boolvar ' + bits;