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Proof step 15 from proof bitsxor-append-same-size (previous step)

# bits0 = # bits2, ∀ bits1, bits, bits2, bits3. # bits3 < # bits0 → # bits3 = # bits2 → bits = [] ∨ bits3 = [] ∨ bits1 = [] ∨ bitsxor(bits + bits3, bits1 + bits2) = bitsxor(bits, bits1) + bitsxor(bits3, bits2) ⊦ # bits2 = 1, bitsxor(bits + bits0.butlast, bits1 + bits2.butlast) = bitsxor(bits, bits1) + bitsxor(bits0.butlast, bits2.butlast), bits = [], bits0 = [], bits1 = [], bits2 = [], # bits1 + # bits2 = 0, # bits + # bits2 = 0

User applied rule apply induction

And obtained the following new goals

Proof step 16:
bits = [] ∨ bits0.butlast = [] ∨ bits1 = [] ∨ bitsxor(bits + bits0.butlast, bits1 + bits2.butlast) = bitsxor(bits, bits1) + bitsxor(bits0.butlast, bits2.butlast), # bits0 = # bits2, ∀ bits1, bits, bits2, bits3. # bits3 < # bits0 → # bits3 = # bits2 → bits = [] ∨ bits3 = [] ∨ bits1 = [] ∨ bitsxor(bits + bits3, bits1 + bits2) = bitsxor(bits, bits1) + bitsxor(bits3, bits2) ⊦ # bits2 = 1, bitsxor(bits + bits0.butlast, bits1 + bits2.butlast) = bitsxor(bits, bits1) + bitsxor(bits0.butlast, bits2.butlast), bits = [], bits0 = [], bits1 = [], bits2 = [], # bits1 + # bits2 = 0, # bits + # bits2 = 0

Rule argument

subst: [bits1, bits, bits2, bits3] → [bits1, bits, bits2.butlast, bits0.butlast]

Additional information

Used simplifier rules: ⊦ 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 + 1 ↔ ¬ n < m;% ⊦ ¬ n < n;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m; Used terms: bits1, bits, bits2.butlast, bits0.butlast