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Proof step 56 from proof bitsor-bi-sgn (previous step)

1 < #(boolvar1 ' + boolvar ') ∧ 1 < #(bits3 + boolvar0 '), ∀ bits, bits1. # bits1 < #(bits3 + boolvar0 ') → true → bits ≠ [] ∧ bits1 ≠ [] → bits2int(bitsor(sgn(bits), bits1)) = bits2int(bitsor(bits, bits1)) ⊦ bits3 = [], bits2int(bitsor((boolvar1 ' + boolvar ').butlast, (bits3 + boolvar0 ').butlast) + ((boolvar1 ' + boolvar ').last ∨ (bits3 + boolvar0 ').last)) = bits2int(bitsor(boolvar1 ' + bits0, bits3) + (boolvar ∨ boolvar0)'), boolvar1 ↔ boolvar, (¬ boolvar1) ∈ bits0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 57:
∀ bits, bits1. # bits1 < #(bits3 + boolvar0 ') → true → bits ≠ [] ∧ bits1 ≠ [] → bits2int(bitsor(sgn(bits), bits1)) = bits2int(bitsor(bits, bits1)) ⊦ boolvar1 ↔ boolvar, bits3 = [], bits2int(bitsor(boolvar1 ', bits3) + (boolvar ∨ boolvar0)') = bits2int(bitsor(boolvar1 ' + bits0, bits3) + (boolvar ∨ boolvar0)'), (¬ boolvar1) ∈ bits0

Rule argument

Additional information

Used simplifier rules: ⊦ # 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;