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Proof step 72 from proof bxor-flip-one-bit (previous step)

int2bits(i) = boolvar0 ' + boolvar ', boolvar0 ∈ boolvar0 ' + boolvar ', boolvar ∈ boolvar0 ' + boolvar ', 0 < i, 2 < i ⊦ i = 0, (sgn(boolvar0 ' + (¬ boolvar)')[# boolvar0 ' + # boolvar ' - 1] ↔ ¬ boolvar) ∧ sgn(boolvar0 ' + (¬ boolvar)')[# boolvar0 ' + # boolvar ' - 1, boolvar] = boolvar0 ' + boolvar '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ # boolvar ' = 0 +1;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ∈ boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ boolvar ∈ bits + bits0 ↔ ¬ ¬(boolvar ∈ bits ∨ boolvar ∈ bits0);