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Proof step 21 from proof bitsand-onein (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
∀ bits0, bits, bits2. # bits2 < # bits1 → true → bits2 = [] ∨ (true ∈ bitsand(bits + false ' + bits0, bits2) ↔ true ∈ bitsand(false ' + bits0, bits2)) ∨ true ∈ bits ⊦ bits0 = [], bits1 = [], true ∈ bitsand(bits + false ' + bits0, bits1) ↔ true ∈ bitsand(false ' + bits0.butlast, bits1.butlast) ∨ bits0.last ∧ bits1.last, # bits1 = 1, true ∈ bits

Rule argument

Additional information

Used simplifier rules: ⊦ 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);% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ∈ boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ boolvar ∈ bits + bits0 ↔ ¬ ¬(boolvar ∈ bits ∨ boolvar ∈ bits0);% ⊦ bits + boolvar = bits + boolvar ';