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Proof step 8 from proof fill-first (previous step)

¬(boolvar ' + bits0 ≠ [] ∧ #(boolvar ' + bits0) < n), fillsgn(n, boolvar ' + bits0) = boolvar0 ' + bits ⊦ boolvar0 ↔ boolvar

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
fillsgn(n, boolvar ' + bits0) = boolvar0 ' + bits, n ≤ 1 + # bits0 ⊦ boolvar0 ↔ boolvar

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ m < n ↔ n ≤ m;% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ' ≠ [];% ⊦ ¬ bits + bits0 ≠ [] ↔ bits = [] ∧ bits0 = [];