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Proof step 41 from proof bitsor-mklist-in (previous step)

1 < #(boolvar0 ' + bits2 + boolvar1 ') ∧ 1 < #(bits3 + boolvar2 '), ∀ bits, boolvar0, boolvar, n, bits1. # bits1 < #(bits3 + boolvar2 ') → true → n = 0 ∨ bits1 = [] ∨ (boolvar ∈ bitsor(mklist(boolvar0, n) + bits, bits1) ↔ boolvar ∈ bitsor(boolvar0 ' + bits, bits1)) ⊦ n = 0, boolvar ∈ bitsor(mklist(boolvar0, n) + bits2, bits3) ∨ (boolvar ↔ boolvar1 ∨ boolvar2) ↔ boolvar ∈ bitsor((boolvar0 ' + bits2 + boolvar1 ').butlast, (bits3 + boolvar2 ').butlast) + ((boolvar0 ' + bits2 + boolvar1 ').last ∨ (bits3 + boolvar2 ').last), bits3 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 42:
∀ bits, boolvar0, boolvar, n, bits1. # bits1 < #(bits3 + boolvar2 ') → true → n = 0 ∨ bits1 = [] ∨ (boolvar ∈ bitsor(mklist(boolvar0, n) + bits, bits1) ↔ boolvar ∈ bitsor(boolvar0 ' + bits, bits1)) ⊦ bits3 = [], n = 0, boolvar ∈ bitsor(mklist(boolvar0, n) + bits2, bits3) ↔ boolvar ∈ bitsor(boolvar0 ' + bits2, bits3), boolvar ↔ boolvar1 ∨ boolvar2

Rule argument

Additional information

Used simplifier rules: ⊦ m + 1 ≠ 0;% ⊦ # bits = 0 ↔ bits = [];% ⊦ m + n = 1 ↔ m = 0 ∧ n = 1 ∨ n = 0 ∧ m = 1;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;%n < n0 ⊦ (* n) ≠ m0 + (* n0);% ⊦ 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;% ⊦ (boolvar ' + bits + boolvar0 ').last ↔ boolvar0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ (bits + boolvar ').butlast = bits;