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Proof step 6 from proof succ-fill (previous step)

¬ n ≤ #(boolvar ' + bits0), false ∈ bits0 ⊦ fillsgn(n, boolvar ' + bits0) +1 = fillsgn(n, boolvar ' + bits0 +1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
1 + # bits0 < n, false ∈ bits0 ⊦ fillsgn(n, boolvar ' + bits0) +1 = fillsgn(n, boolvar ' + bits0 +1)

Rule argument

Additional information

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