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Proof step 3 from proof fill-sgn-03 (previous step)

bits = boolvar + bits0, # bits ≤ n + 1 ⊦ bits = [], fillsgn(n + 1, sgn(bits0)).rest = fillsgn(n, bits0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
# bits0 ≤ n ⊦ fillsgn(n + 1, sgn(bits0)).rest = fillsgn(n, bits0)

Rule argument

Additional information

Used simplifier rules: ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n ≤ m + n0 ↔ n ≤ n0;% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ' + bits ≠ [];% ⊦ boolvar + bits = boolvar ' + bits;