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Proof step 163 from proof byte-law-and-one (previous step)

# bits2 + # bits1 = 8, int2bits(j) = (bits0 + bits2) + bits1, # bits1 = # int2bits(i), 7 = # fillsgn(7, int2bits(i)), 8 ≤ # bitsand(bits2 + bits1, fillsgn(8, int2bits(i))), 8 ≤ # bits0 + # bitsand(bits2 + bits1, fillsgn(8, int2bits(i))), # bits1 < 8, i < 0, # bits1 < # bitsand(bits2 + bits1, fillsgn(8, int2bits(i))), # bits1 < # bits0 + # bitsand(bits2 + bits1, fillsgn(8, int2bits(i))) ⊦ bitsand(bits2 + bits1, true ' + fillsgn(7, int2bits(i))) = [], lastn(8, fillsgn(8, bitsand((bits0 + bits2) + bits1, int2bits(i)))) = lastn(8, bitsand(bits2 + bits1, true ' + fillsgn(7, int2bits(i)))), bitsand(bits2 + bits1, fillsgn(8, int2bits(i))) = [], bits2 = []

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

seq: ⊦ # bits1 ≤ # bits0 ∧ bits1 ≠ [] ∧ bits1.first → bitsand(bits + bits0, bits1) = bits + bitsand(bits0, bits1) subst: [bits1, bits0, bits] → [int2bits(i), bits1, bits0 + bits2] rotate: F spec: intops inst: name: bitsand-app-true

Additional information

Used simplifier rules: ⊦ ¬ m ≤ n ↔ n < m;% ⊦ int2bits(j) ≠ [];% ⊦ j < 0 → int2bits(j).first;% ⊦ n < n0 ∧ n0 < n1 → n +1 < n1; (forward)% ⊦ m < n ∧ m = n0 → n0 < n; (forward)% ⊦ m < n ∧ n = n0 → m < n0; (forward)% ⊦ m < n → n ≠ 0; (forward)% ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward)% ⊦ m + n = n + m;% ⊦ # bits = 0 ↔ bits = [];% ⊦ m +1 = m + 1;% ⊦ (m + n0) + n1 = m + n0 + n1;%n < n0 ⊦ m + (* n) < (* n0) ↔ m < n0 - n;% ⊦ m < n → ¬ n < m; Used terms: int2bits(i), bits1, bits0 + bits2