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Proof step 275 from proof byte-law-and-one (previous step)

int2bits(j) = bits2 + bits1, # bits1 = 8, int2bits(i) = bits0 + bits, # bits = 8, 8 ≤ # bitsand(bits0, bits2) + # bitsand(bits1, bits), 8 ≤ # bitsand(bits1, bits) ⊦ # bitsand(bits0, bits2) + # bitsand(bits1, bits) = 0, lastn(8, bitsand(bits1, bits)) = lastn(8, fillsgn(8, bitsand(bits2 + bits1, bits))), bits2 + bits1 = [], bits2 = [], bits0 = [], bits1 = [], bits = [], bits.first

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

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

Additional information

Used simplifier rules: ⊦ n ≤ # bits → fillsgn(n, bits) = bits;%⊦ ¬ bits0.first ∧ # bits0 ≤ # bits ∧ bits0 ≠ [] → # bitsand(bits, bits0) = # bits0;% ⊦ m + n = n + m;% ⊦ ¬ 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)% ⊦ m < n → n ≠ 0; (forward)% ⊦ m < n → n ≠ 0;% ⊦ m < n → ¬ n < m; Used terms: bits, bits1, bits2