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Proof step 266 from proof byte-law-and-one (previous step)

# bits = 8, int2bits(i) = bits0 + bits, # bits1 = 8, int2bits(j) = bits2 + bits1, 8 ≤ # bitsand(bits0, bits2) + # bitsand(bits, bits1), # bitsand(bits, bits1) < 8 ⊦ bits1 = [], bits0 = [], bits2 = [], bits = [], bits2 + bits1 = [], lastn(8, bitsand(bits0, bits2) + bitsand(bits, bits1)) = lastn(8, fillsgn(8, bitsand(bits2 + bits1, bits))), bits1.first

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 267:
⊦ ¬ bits0.first ∧ # bits0 ≤ # bits ∧ bits0 ≠ [] → # bitsand(bits, bits0) = # bits0

Rule argument

seq: ⊦ ¬ bits0.first ∧ # bits0 ≤ # bits ∧ bits0 ≠ [] → # bitsand(bits, bits0) = # bits0 subst: [bits0, bits] → [bits1, bits] rotate: F spec: intops inst: name: bitsand-len-false

Additional information

Used simplifier rules: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward)%0 < n ⊦ (* n) ≤ m → m ≠ 0; (forward)% ⊦ ¬ m ≤ n ↔ n < m;% ⊦ m < n ∧ n ≤ n0 → m < n0; (forward)% ⊦ m < n ∧ m = n0 → n0 < n; (forward)% ⊦ m < n ∧ n = n0 → m < n0; (forward)% ⊦ n < n0 ∧ n0 < n1 → n +1 < n1; (forward)% ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward)% ⊦ m < n → ¬ n < m; Used terms: bits1, bits