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Proof step 11 from proof bitsand-idempotent (previous step)

bits = bits1 + boolvar ', ∀ bits0. # bits0 < # bits → true → bits0 = [] ∨ bitsand(bits0, bits0) = bits0 ⊦ bits = [], bitsand(bits1, bits1) + boolvar ' = bits, # bits = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
∀ bits0. # bits0 < #(bits1 + boolvar ') → true → bits0 = [] ∨ bitsand(bits0, bits0) = bits0 ⊦ bitsand(bits1, bits1) = bits1, bits1 = []

Rule argument

Additional information

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