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Proof step 3 from proof sgn-base-05 (previous step)

bits = boolvar + bits0, boolvar ↔ ¬ bits0.first ⊦ sgn(bits) = bits, # bits = 1, bits = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ sgn((¬ bits0.first)' + bits0) = (¬ bits0.first)' + bits0, bits0 = []

Rule argument

Additional information

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