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Proof step 13 from proof add-ones-grlen-02 (previous step)

# 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

User applied rule apply induction

And obtained the following new goals

Proof step 14:
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

Rule argument

subst: [bits, bits1] → [bits.butlast, bits0.butlast]

Additional information

Used simplifier rules: ⊦ ¬ n < n;% ⊦ m < n + 1 ↔ ¬ n < m;%n0 ≤ n ⊦ n - n0 < m ↔ n < m + n0;% ⊦ bits ≠ [] → # bits.butlast = # bits - 1;% ⊦ # bits = 0 ↔ bits = [];% ⊦ m < 1 ↔ m = 0;% ⊦ ¬ m ≤ n ↔ n < m;%n0 ≤ n ⊦ n - n0 ≤ m ↔ n ≤ m + n0;% ⊦ m < n ∧ n ≤ n0 → m < n0; (forward)% ⊦ m < n → n ≠ 0; (forward)% ⊦ ¬ n + (m - n) < m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m; Used terms: bits.butlast, bits0.butlast