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Proof step 12 from proof bitsor-bi-ones (previous step)

∃ bits0, bits1. bits = bits0 + bits1 ∧ # bits1 = # bits2, 1 + # bits2 < # bits ⊦ bits2int(bitsor(bits, true ' + bits2) + bits1) = bits2int(true ' + bits1), bits = [], false ∈ bits

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
# bits0 = # bits2 ⊦ bits3 + bits0 = [], bits2int(bitsor(bits3 + bits0, true ' + bits2) + bits1) = bits2int(true ' + bits1), # bits3 = 1, bits3 = [], false ∈ bits3, false ∈ bits0

Rule argument

Additional information

Used simplifier rules: ⊦ 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;