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Proof step 18 from proof bitsxor-zeros (previous step)

bits = bits2 + boolvar ', # bits0 ≤ # bits, ∀ bits, bits1. # bits1 < # bits0 → # bits1 ≤ # bits → bits = [] ∨ bits1 = [] ∨ bitsxor(bits, bits1) = bits ∨ true ∈ bits1 ⊦ bits0 = [], bits = [], bitsxor(bits2, bits0.butlast) + (¬(boolvar ↔ bits0.last))' = bits, # bits = 1, # bits0 = 1, true ∈ bits0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
# bits0 ≤ # bits2 + 1, ∀ bits, bits1. # bits1 < # bits0 → # bits1 ≤ # bits → bits = [] ∨ bits1 = [] ∨ bitsxor(bits, bits1) = bits ∨ true ∈ bits1 ⊦ # bits0 = 1, bits0 = [], bitsxor(bits2, bits0.butlast) + (¬(boolvar ↔ bits0.last))' = bits2 + boolvar ', bits2 = [], true ∈ bits0

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;% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + boolvar ' ≠ [];