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Proof step 14 from proof add-ones-grlen-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: bits ≠ [] ⊦ boolvar ∈ bits.butlast → boolvar ∈ bits; (forward)% ⊦ # boolvar ' = 0 +1;% ⊦ bits ≠ [] → (bits.butlast = [] ↔ bits = bits.first ');