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Proof step 8 from proof bitsand-len-false (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
# bits1 < # bits, ∀ bits, bits0. # bits0 < # bits1 → # bits0 < # bits → # bitsand(bits, false ' + bits0) = 1 + # bits0 ∨ bits = [] ⊦ # bits = 1, bits1 = [], bits = [], # bitsand(bits.butlast, false ' + bits1.butlast) = # bits1

Rule argument

Additional information

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