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Proof step 70 from proof and-sgn (previous step)

1 < #(sgn(bits0) + boolvar0 ') ∧ 1 < #(bits2 + boolvar '), (¬ boolvar0) ∈ bits0, ∀ bits1, bits. # bits1 < #(bits2 + boolvar ') → true → bits = [] ∨ bits1 = [] ∨ sgn(bitsand(sgn(bits), bits1)) = sgn(bitsand(bits, bits1)) ⊦ bits2 + boolvar ' = (bits2 + boolvar ').first ', bits0 = [], sgn(bitsand((sgn(bits0) + boolvar0 ').butlast, (bits2 + boolvar ').butlast) + ((sgn(bits0) + boolvar0 ').last ∧ (bits2 + boolvar ').last)) = sgn(bitsand(bits0, bits2) + (boolvar0 ∧ boolvar)'), bits0 + boolvar0 ' = bits0.first '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 71:
(¬ boolvar0) ∈ bits0, ∀ bits1, bits. # bits1 < #(bits2 + boolvar ') → true → bits = [] ∨ bits1 = [] ∨ sgn(bitsand(sgn(bits), bits1)) = sgn(bitsand(bits, bits1)) ⊦ bits0 = [], sgn(bitsand(sgn(bits0), bits2) + (boolvar0 ∧ boolvar)') = sgn(bitsand(bits0, bits2) + (boolvar0 ∧ boolvar)'), bits2 = [], bits0 + boolvar0 ' = bits0.first ', bits2 + boolvar ' = bits2.first '

Rule argument

Additional information

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