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Proof step 979 from proof bitsand-cast (previous step)

# bits = # bits1 ⊦ bits3 = [], bits1 = [], bits3 + bits = [], bits1 = lastn(# bits1, fillsgn(# bits1, bitsand(bits1, bits3 + bits))), true ∈ bits1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 980:
⊦ ¬ true ∈ bits ∧ (# bits ≤ # bits0 ∨ bits0.first) ∧ bits ≠ [] ∧ bits0 ≠ [] → bitsand(bits, bits0) = bits
Proof step 981:
¬ true ∈ bits1 ∧ (# bits1 ≤ #(bits3 + bits) ∨ (bits3 + bits).first) ∧ bits1 ≠ [] ∧ bits3 + bits ≠ [], # bits = # bits1 ⊦ bits3 = [], bits1 = [], bits3 + bits = [], bits1 = lastn(# bits1, fillsgn(# bits1, bits1)), true ∈ bits1

Rule argument

name: bitsand-zeros rotate: F subst: [bits0, bits] → [bits3 + bits, bits1]

Additional information

Used simplifier rules: ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ ¬ m ≤ n ↔ n < m;% ⊦ bits ≠ [] → ((bits + bits0).first ↔ bits.first);% ⊦ m < n ∧ n = n0 → m < n0; (forward)% ⊦ m < n → n ≠ 0; (forward)% ⊦ # bits = 0 ↔ bits = [];% ⊦ ¬ m + n < m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m; Used terms: bits3 + bits, bits1