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Proof step 269 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(bits, bits1), 8 ≤ # bitsand(bits, bits1) ⊦ # bitsand(bits0, bits2) + # bitsand(bits, bits1) = 0, lastn(8, bitsand(bits, bits1)) = lastn(8, fillsgn(8, bitsand(bits2 + bits1, bits))), bits2 + bits1 = [], bits = [], bits2 = [], bits0 = [], bitsand(bits, bits1) = []

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 270:
⊦ bits ≠ [] ∧ bits0 ≠ [] → bitsand(bits, bits0) = bitsand(bits0, bits)
Proof step 271:
bits ≠ [] ∧ bits1 ≠ [], 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 = [], bits = [], bits2 = [], bits0 = [], bitsand(bits1, bits) = []

Rule argument

seq: ⊦ bits ≠ [] ∧ bits0 ≠ [] → bitsand(bits, bits0) = bitsand(bits0, bits) subst: [bits0, bits] → [bits1, bits] rotate: F spec: intops inst: name: and-com

Additional information

Used simplifier rules: ⊦ # [] = 0; Used terms: bits1, bits