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Proof step 207 from proof byte-law-and-one (previous step)

# bits = 8, int2bits(i) = bits0 + bits, # bitsand(bits, int2bits(j)) < 8, # int2bits(j) < 8, j < 0 ⊦ lastn(8, fillsgn(8, bits0 + bitsand(bits, int2bits(j)))) = lastn(8, bitsand(bits, int2bits(j)).first ' + fillsgn(7, bitsand(bits, int2bits(j)))), bitsand(bits, int2bits(j)) = []

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 208:
⊦ bits0.first ∧ # bits0 ≤ # bits ∧ bits0 ≠ [] → # bitsand(bits, bits0) = # bits

Rule argument

seq: ⊦ bits0.first ∧ # bits0 ≤ # bits ∧ bits0 ≠ [] → # bitsand(bits, bits0) = # bits subst: [bits0, bits] → [int2bits(j), bits] rotate: F spec: intops inst: name: bitsand-len-true

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (forward)% ⊦ ¬ m ≤ n ↔ n < m;% ⊦ int2bits(j) ≠ [];% ⊦ j < 0 → int2bits(j).first;% ⊦ n < n0 ∧ n0 < n1 → n +1 < n1; (forward)% ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward)% ⊦ m < n ∧ m = n0 → n0 < n; (forward)% ⊦ m < n ∧ n = n0 → m < n0; (forward)% ⊦ m +1 = m + 1;%n < n0 ⊦ m + (* n) < (* n0) ↔ m < n0 - n;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m;% ⊦ m < n → ¬ n < m; Used terms: int2bits(j), bits