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Proof step 54 from proof and-insgn (previous step)

1 < #(sgn(bits0) + boolvar1 ') ∧ 1 < #(bits2 + boolvar0 '), (¬ boolvar1) ∈ bits0, ∀ bits1, bits, boolvar. # bits1 < #(bits2 + boolvar0 ') → true → bits = [] ∨ bits1 = [] ∨ (boolvar ∈ bitsand(sgn(bits), bits1) ↔ boolvar ∈ bitsand(bits, bits1)) ⊦ boolvar ∈ bitsand((sgn(bits0) + boolvar1 ').butlast, (bits2 + boolvar0 ').butlast) + ((sgn(bits0) + boolvar1 ').last ∧ (bits2 + boolvar0 ').last) ↔ (boolvar ↔ boolvar1 ∧ boolvar0) ∨ boolvar ∈ bitsand(bits0, bits2), bits0 + boolvar1 ' = (bits0 + boolvar1 ').first ', bits2 + boolvar0 ' = (bits2 + boolvar0 ').first '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 55:
(¬ boolvar1) ∈ bits0, ∀ bits1, bits, boolvar. # bits1 < #(bits2 + boolvar0 ') → true → bits = [] ∨ bits1 = [] ∨ (boolvar ∈ bitsand(sgn(bits), bits1) ↔ boolvar ∈ bitsand(bits, bits1)) ⊦ (boolvar ↔ boolvar1 ∧ boolvar0) ∨ boolvar ∈ bitsand(sgn(bits0), bits2) ↔ (boolvar ↔ boolvar1 ∧ boolvar0) ∨ boolvar ∈ bitsand(bits0, bits2), bits0 = [], bits0 + boolvar1 ' = bits0.first ', bits2 = [], bits2 + boolvar0 ' = 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;% ⊦ boolvar ∈ boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ boolvar ∈ bits + bits0 ↔ ¬ ¬(boolvar ∈ bits ∨ boolvar ∈ bits0);% ⊦ bits + boolvar = bits + boolvar ';% ⊦ (bits + boolvar ').last ↔ boolvar;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ (bits + boolvar ').butlast = bits;