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Proof step 20 from proof bitsor-mklist-in (previous step)

1 < #(mklist(boolvar0, n) + bits) ∧ 1 < # bits0, ∀ bits, boolvar0, boolvar, n, bits1. # bits1 < # bits0 → true → n = 0 ∨ bits1 = [] ∨ (boolvar ∈ bitsor(mklist(boolvar0, n) + bits, bits1) ↔ boolvar ∈ bitsor(boolvar0 ' + bits, bits1)) ⊦ boolvar ∈ bitsor((mklist(boolvar0, n) + bits).butlast, bits0.butlast) + ((mklist(boolvar0, n) + bits).last ∨ bits0.last) ↔ boolvar ∈ bitsor(boolvar0 ' + bits, bits0), bits0 = [], n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
∀ bits, boolvar0, boolvar, n, bits1. # bits1 < # bits0 → true → n = 0 ∨ bits1 = [] ∨ (boolvar ∈ bitsor(mklist(boolvar0, n) + bits, bits1) ↔ boolvar ∈ bitsor(boolvar0 ' + bits, bits1)) ⊦ bits0 = [], n = 0, boolvar ∈ bitsor((mklist(boolvar0, n) + bits).butlast, bits0.butlast) ∨ (boolvar ↔ (mklist(boolvar0, n) + bits).last ∨ bits0.last) ↔ boolvar ∈ bitsor(boolvar0 ' + bits, bits0), n + # bits = 0, # bits0 = 1, bits = [] ∧ n = 1

Rule argument

Additional information

Used simplifier rules: ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ # bits = 0 ↔ bits = [];% ⊦ m + n = 1 ↔ m = 0 ∧ n = 1 ∨ n = 0 ∧ m = 1;% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1);% ⊦ # mklist(boolvar, n) = n;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ∈ boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ boolvar ∈ bits + bits0 ↔ ¬ ¬(boolvar ∈ bits ∨ boolvar ∈ bits0);% ⊦ bits + boolvar = bits + boolvar ';