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Proof step 985 from proof bitsand-cast (previous step)

# bits = # bits1 ⊦ bits2 = [], bits3 = [], bits1 = [], bits2 + bits1 = [], bits3 + bits = [], bits1 = lastn(# bits1, fillsgn(# bits1, bitsand(bits2, bits3) + bitsand(bits1, bits))), true ∈ bits1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 986:
⊦ ¬ true ∈ bits ∧ (# bits ≤ # bits0 ∨ bits0.first) ∧ bits ≠ [] ∧ bits0 ≠ [] → bitsand(bits, bits0) = bits
Proof step 987:
¬ true ∈ bits1 ∧ (# bits1 ≤ # bits ∨ bits.first) ∧ bits1 ≠ [] ∧ bits ≠ [], # bits = # bits1 ⊦ bits2 = [], bits3 = [], bits1 = [], bits2 + bits1 = [], bits3 + bits = [], bits1 = lastn(# bits1, fillsgn(# bits1, bitsand(bits2, bits3) + bits1)), true ∈ bits1

Rule argument

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

Additional information

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