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Proof step 368 from proof bitsand-cast (previous step)

true ∈ bits1 ∧ false ∈ bits1 → (∃ boolvar, n, boolvar0, bits0. bits1 = mklist(boolvar, n) + (boolvar0 + bits0) ∧ sgn(bits1) = boolvar + boolvar0 + bits0 ∧ ¬(boolvar ↔ boolvar0) ∧ n ≠ 0) ⊦ bits2 + bits1 = [], bits0 = [], lastn(# bits1, fillsgn(# bits1, bitsand(sgn(bits1), bits0))) = lastn(# bits1, fillsgn(# bits1, bitsand(bits2 + bits1, bits0)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 369:
true ∈ bits1 ∧ false ∈ bits1 → bits1 = mklist(boolvar, n) + boolvar0 ' + bits ∧ sgn(mklist(boolvar, n) + boolvar0 ' + bits) = boolvar ' + boolvar0 ' + bits ∧ ¬(boolvar ↔ boolvar0) ∧ n ≠ 0 ⊦ bits2 + bits1 = [], bits0 = [], lastn(# bits1, fillsgn(# bits1, bitsand(sgn(bits1), bits0))) = lastn(# bits1, fillsgn(# bits1, bitsand(bits2 + bits1, bits0)))

Rule argument

Additional information

Used simplifier rules: ⊦ boolvar + bits = boolvar ' + bits;