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Proof step 22 from proof and-true (previous step)

1 < # bits ∧ 1 < # bits0, bitsand(bits.butlast, bits0.butlast) + (bits.last ∧ bits0.last) = true ' ⊦ bits0 = [], bits = [], bits = bitsand(bits.butlast, bits0.butlast) + (bits.last ∧ bits0.last) ∧ bits0 = bitsand(bits.butlast, bits0.butlast) + (bits.last ∧ bits0.last)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 23:
bitsand(bits.butlast, bits0.butlast) = [], bits0.last, bits.last ⊦ bits0 = bits0.first ', bits = bits.first ', bits = [], bits0 = [], bits = true '

Rule argument

Additional information

Used simplifier rules: ⊦ boolvar ' = boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ boolvar '.first ↔ boolvar;% ⊦ boolvar ' ≠ [];% ⊦ boolvar '.butlast = [];% ⊦ [] + bits = bits;% ⊦ bits + [] = bits;% ⊦ boolvar + bits = boolvar ' + bits;% ⊦ # bits = 1 ↔ bits = bits.first + [];% ⊦ # bits = 0 ↔ bits = [];% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1);% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ' = bits + boolvar0 ' ↔ (boolvar ↔ boolvar0) ∧ bits = [];% ⊦ bits + boolvar = bits + boolvar ';