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Proof step 245 from proof byte-law-and-one (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 246:
⊦ bits ≠ [] ∧ bits0 ≠ [] → bitsand(bits, bits0) = bitsand(bits0, bits)
Proof step 247:
bits ≠ [] ∧ bits2 + bits1 ≠ [], int2bits(j) = bits2 + bits1, # bits1 = 8, int2bits(i) = bits0 + bits, # bits = 8 ⊦ bits2 = [], bits0 = [], bitsand(bits2 + bits1, bits) = [], lastn(8, fillsgn(8, bitsand(bits0 + bits, bits2 + bits1))) = lastn(8, fillsgn(8, bitsand(bits2 + bits1, bits)))

Rule argument

seq: ⊦ bits ≠ [] ∧ bits0 ≠ [] → bitsand(bits, bits0) = bitsand(bits0, bits) subst: [bits0, bits] → [bits2 + bits1, bits] rotate: F spec: intops inst: name: and-com

Additional information

Used simplifier rules: ⊦ ¬ bits + bits0 ≠ [] ↔ bits = [] ∧ bits0 = [];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ # [] = 0; Used terms: bits2 + bits1, bits