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Proof step 6 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, bits0) ↔ boolvar ∈ bitsor(boolvar0 ' + bits, bits0), bits0 = [], n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
∀ 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, bits0) ↔ boolvar ∈ bitsor(boolvar0 ' + bits, bits0), bits0 = [], n = 0, (n ≠ 1 ∨ bits ≠ []) ∧ # bits0 ≠ 1

Rule argument

Additional information

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