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Proof step 8 from proof and-size-one (previous step)

bits.first, # bits ≤ # bits1, ∀ bits2, bits1, bits0. # bits2 < # bits → # bits2 ≤ # bits1 ∧ bits2.first → bits2 = [] ∨ bitsand(bits2, bits0 + bits1) = bits0 + bitsand(bits2, bits1) ⊦ bits1 = [], bitsand(bits, bits0 + bits1) = bits0 + bitsand(bits, bits1), bits = [], (bits0 ≠ [] ∨ bits1 ≠ []) ∧ # bits0 + # bits1 ≠ 1 ∧ bits ≠ true '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
bits.first, # bits ≤ # bits1, ∀ bits2, bits1, bits0. # bits2 < # bits → # bits2 ≤ # bits1 ∧ bits2.first → bits2 = [] ∨ bitsand(bits2, bits0 + bits1) = bits0 + bitsand(bits2, bits1) ⊦ bits1 = [], bitsand(bits, bits0 + bits1) = bits0 + bitsand(bits, bits1), bits = [], # bits0 + # bits1 ≠ 1 ∧ bits ≠ true '

Rule argument

Additional information

Used simplifier rules: and-onel-01: ⊦ bits ≠ [] → bitsand(true ', bits) = bits;% ⊦ boolvar '.first ↔ boolvar;% ⊦ # boolvar ' = 0 +1;% ⊦ boolvar ' ≠ [];% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + [] = bits;% ⊦ boolvar + bits = boolvar ' + bits;% ⊦ # bits = 1 ↔ bits = bits.first + [];% ⊦ # bits = 0 ↔ bits = [];% ⊦ m + n = 1 ↔ m = 0 ∧ n = 1 ∨ n = 0 ∧ m = 1;