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Proof step 25 from proof bitsor-sgn-in (previous step)

∃ bits0, bits2. bits = bits0 + bits2 ∧ # bits2 = # bits1, 1 + # bits1 < # bits, true ∈ bits ⊦ boolvar ↔ boolvar ∈ bitsor(bits, true ' + bits1), bits = [], false ∈ bits

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 26:
# bits0 = # bits1, true ∈ bits2 ⊦ bits2 + bits0 = [], boolvar ↔ boolvar ∈ bitsor(bits2 + bits0, true ' + bits1), # bits2 = 1, bits2 = [], false ∈ bits2, false ∈ bits0

Rule argument

Additional information

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