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Proof step 15 from proof and-ones (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
bitsand(true ' + bits2.butlast, false ' + bits0.butlast) = false ' + bits0.butlast, bits0.last, ∀ bits1, bits0. # bits1 < #(true + bits2) → true → bits1 = [] ∨ bitsand(bits1, false ' + bits0) = false ' + bits0 ∨ false ∈ bits1 ∨ false ∈ bits0 ⊦ bits0 = [], bits2 = [], false ∈ bits2, false ∈ bits0, false ∈ bits0.butlast, false ∈ bits2.butlast, bits2.last

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ boolvar ∈ bits → (boolvar ∈ boolvar0 ' + bits ↔ boolvar ↔ boolvar0);% ⊦ boolvar ' + bits = bits0 → boolvar ∈ bits0; (forward)%bits ≠ [] ⊦ boolvar ∈ bits.butlast → boolvar ∈ bits; (forward)% ⊦ boolvar ∈ boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ boolvar ∈ bits + bits0 ↔ ¬ ¬(boolvar ∈ bits ∨ boolvar ∈ bits0);% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits ≠ [] → (bits.butlast = [] ↔ bits = bits.first ');% ⊦ boolvar ' ≠ [];% ⊦ ¬ bits + bits0 ≠ [] ↔ bits = [] ∧ bits0 = [];