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Proof step 4 from proof bitsand-app (previous step)

# bits = # bits1, ∀ bits2, bits0, bits, bits3. # bits3 < # bits1 → # bits = # bits3 → bits0 = [] ∨ bits2 = [] ∨ bits = [] ∨ bitsand(bits0 + bits, bits2 + bits3) = bitsand(bits0, bits2) + bitsand(bits, bits3) ⊦ bitsand(bits0 + bits, bits2 + bits1) = bitsand(bits0, bits2) + bitsand(bits, bits1), bits = [], bits2 = [], bits0 = []

User applied rule insert rewrite lemma

And obtained the following new goals

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

Rule argument

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

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: bits2 + bits1, bits0 + bits