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Proof step 6 from proof sgn-length (previous step)

¬(bits0 ≠ [] ∧ (bits0.first ↔ boolvar)), 1 + # bits0 < # sgn(boolvar ' + bits0), ¬ # sgn(bits0) ≤ # bits0 → sgn(bits0) = [] ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
1 + # bits0 < # sgn(boolvar ' + bits0), # sgn(bits0) ≤ # bits0 ⊦ bits0.first ↔ boolvar, bits0 = []

Rule argument

Additional information

Used simplifier rules: sgn-one: ⊦ sgn(boolvar ') = boolvar ';%sgn-not-empty: ⊦ bits ≠ [] → sgn(bits) ≠ [];% ⊦ # boolvar ' = 0 +1;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + [] = bits;% ⊦ # [] = 0;