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Proof step 254 from proof byte-law-and-one (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 255:
⊦ bits0.first ∧ # bits0 ≤ # bits ∧ bits0 ≠ [] → # bitsand(bits, bits0) = # bits
Proof step 256:
¬(bits1.first ∧ # bits1 ≤ # bits ∧ bits1 ≠ []), int2bits(j) = bits2 + bits1, # bits1 = 8, int2bits(i) = bits0 + bits, # bits = 8, # bitsand(bits0, bits2) + # bitsand(bits, bits1) < 8 ⊦ lastn(8, fillsgn(8, bitsand(bits0, bits2) + bitsand(bits, bits1))) = lastn(8, fillsgn(8, bitsand(bits2 + bits1, bits))), bits2 + bits1 = [], bits = [], bits2 = [], bits0 = []

Rule argument

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

Additional information

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