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Proof step 14 from proof bitsor-first (previous step)

1 < #(true ' + bits) ∧ 1 < # bits0, bitsor((true ' + bits).butlast, bits0.butlast) + ((true ' + bits).last ∨ bits0.last) = false ' + bits1, ∀ bits1, bits, bits2. # bits2 < # bits0 → bitsor(true ' + bits, bits2) = false ' + bits1 → bits2 = [] ⊦ bits0 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
bitsor(true ' + bits.butlast, bits0.butlast) + (bits.last ∨ bits0.last)' = false ' + bits1, ∀ bits1, bits, bits2. # bits2 < # bits0 → bitsor(true ' + bits, bits2) = false ' + bits1 → bits2 = [] ⊦ # bits0 = 1, bits = [], bits0 = [], (bits.last ∨ bits0.last) ∧ ¬ true ∈ bits1, ¬ false ∈ bitsor(true ' + bits.butlast, bits0.butlast) ∧ (bits.last ∨ bits0.last)

Rule argument

Additional information

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