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Proof step 4 from proof sgn-length-02 (previous step)

¬(bits ≠ [] ∧ (bits.first ↔ boolvar)), 2 + # bits ≤ # sgn(boolvar ' + bits) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
2 + # bits ≤ # sgn(boolvar ' + bits) ⊦ bits.first ↔ boolvar, bits = []

Rule argument

Additional information

Used simplifier rules: sgn-one: ⊦ sgn(boolvar ') = boolvar ';% ⊦ # boolvar ' = 0 +1;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + [] = bits;% ⊦ # [] = 0;