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Proof step 4 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) ⊦ bitsxor(bits + bits0, bits1 + bits2) = bitsxor(bits, bits1) + bitsxor(bits0, bits2), bits1 = [], bits0 = [], bits = []

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ 1 < # bits ∧ 1 < # bits0 → bitsxor(bits, bits0) = bitsxor(bits.butlast, bits0.butlast) + (¬(bits.last ↔ bits0.last))
Proof step 6:
1 < #(bits + bits0) ∧ 1 < #(bits1 + bits2), # bits0 = # bits2, ∀ bits1, bits, bits2, bits3. # bits3 < # bits0 → # bits3 = # bits2 → bits = [] ∨ bits3 = [] ∨ bits1 = [] ∨ bitsxor(bits + bits3, bits1 + bits2) = bitsxor(bits, bits1) + bitsxor(bits3, bits2) ⊦ bitsxor((bits + bits0).butlast, (bits1 + bits2).butlast) + (¬((bits + bits0).last ↔ (bits1 + bits2).last)) = bitsxor(bits, bits1) + bitsxor(bits0, bits2), bits1 = [], bits0 = [], bits = []

Rule argument

name: bitsxor-rec-01 rotate: F subst: [bits0, bits] → [bits1 + bits2, bits + bits0]

Additional information

Used simplifier rules: ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1);% ⊦ m + n = 0 ↔ m = 0 ∧ n = 0;% ⊦ # bits = 0 ↔ bits = [];% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m;% ⊦ # bits + # bits0 = 1 ∧ bits ≠ [] → # bits = 1 ∧ bits0 = []; (forward)% ⊦ m + n = 1 ↔ m = 0 ∧ n = 1 ∨ n = 0 ∧ m = 1;% ⊦ # [] = 0;% ⊦ 0 + m = m; Used terms: bits1 + bits2, bits + bits0