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Proof step 29 from proof bitsand-fillsgn (previous step)

∀ n, bits, bits1. # bits1 < #(boolvar + bits2) → true → bits = [] ∨ bits1 = [] ∨ fillsgn(n, bitsand(fillsgn(n, bits), bits1)) = fillsgn(n, bitsand(bits, bits1)) ⊦ n = 0, fillsgn(n, bitsand(mklist(false, n), boolvar ' + bits2)) = mklist(false, n), ¬ n ≤ 1 + # bits2 ∧ ¬ boolvar

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 30:
⊦ ¬ true ∈ bits ∧ (# bits ≤ # bits0 ∨ bits0.first) ∧ bits ≠ [] ∧ bits0 ≠ [] → bitsand(bits, bits0) = bits

Rule argument

name: bitsand-zeros rotate: F subst: [bits0, bits] → [boolvar ' + bits2, mklist(false, n)]

Additional information

Used simplifier rules: ⊦ mklist(boolvar, m) = mklist(boolvar, n) ↔ m = n;% ⊦ fillsgn(n, mklist(boolvar, n)) = mklist(boolvar, n);% ⊦ ¬ true ∈ mklist(false, n);% ⊦ # mklist(boolvar, n) = n;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ # boolvar ' = 0 +1;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ ¬ m ≤ n ↔ n < m;% ⊦ (boolvar ' + bits).first ↔ boolvar;% ⊦ mklist(boolvar, m) = [] ↔ m = 0;% ⊦ ¬ bits + bits0 ≠ [] ↔ bits = [] ∧ bits0 = [];% ⊦ boolvar ' ≠ [];% ⊦ m < n → n ≠ 0;% ⊦ m + n = n + m;% ⊦ m < n → (n ≤ m ↔ false); Used terms: boolvar ' + bits2, mklist(false, n)