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Proof step 6 from proof bitsand-app (previous step)

1 < #(bits0 + bits) ∧ 1 < #(bits2 + bits1), # bits = # bits1, ∀ bits2, bits0, bits, bits3. # bits3 < # bits1 → # bits = # bits3 → bits0 = [] ∨ bits2 = [] ∨ bits = [] ∨ bitsand(bits0 + bits, bits2 + bits3) = bitsand(bits0, bits2) + bitsand(bits, bits3) ⊦ bitsand((bits0 + bits).butlast, (bits2 + bits1).butlast) + ((bits0 + bits).last ∧ (bits2 + bits1).last) = bitsand(bits0, bits2) + bitsand(bits, bits1), bits = [], bits2 = [], bits0 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
# bits = # bits1, ∀ bits2, bits0, bits, bits3. # bits3 < # bits1 → # bits = # bits3 → bits0 = [] ∨ bits2 = [] ∨ bits = [] ∨ bitsand(bits0 + bits, bits2 + bits3) = bitsand(bits0, bits2) + bitsand(bits, bits3) ⊦ # bits0 + # bits1 = 0, # bits2 + # bits1 = 0, bits1 = [], bits = [], bits2 = [], bits0 = [], bitsand(bits0 + bits.butlast, bits2 + bits1.butlast) + (bits.last ∧ bits1.last)' = bitsand(bits0, bits2) + bitsand(bits, bits1)

Rule argument

Additional information

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