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Proof step 1 from proof byte-law-righthift-not (conclusion)

⊦ (i = 128 ∧ j = 1 → i2b(i i>> j) ≠ i2b(b2i(i2b(i)) i>> j)) ∧ (i = 128 ∧ j = 1 → i2b(i i>>> j) ≠ i2b(b2i(i2b(i)) i>>> j))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ ( i = 128 ∧ j = 1 → i2b ((# int2bits(128) ≤ i→n(bits2int(bitsand(int2bits(1), int2bits(31)))) ⊃ 0 ;bits2int(sublist(0, # int2bits(128) - i→n(bits2int(bitsand(int2bits(1), int2bits(31)))), int2bits(128))) )) ≠ i2b ((# int2bits(~128) ≤ i→n(bits2int(bitsand(int2bits(1), int2bits(31)))) ⊃ ~1 ;bits2int(sublist(0, # int2bits(~128) - i→n(bits2int(bitsand(int2bits(1), int2bits(31)))), int2bits(~128))) ))) ∧ (i = 128 ∧ j = 1 → i2b(128 i>>> 1) ≠ i2b(~128 i>>> 1))

Rule argument

Additional information

Used simplifier rules: bib-pos-high: ⊦ 0 ≤ i ∧ 127 < i % 256 → b2i(i2b(i)) = i % 256 - 256;% ⊦ band(i, j) = band(j, i);% ⊦ band(i, band(i0, j)) = band(band(i, i0), j);% ⊦ band((* i), (* j)) = bits2int(bitsand(int2bits(i), int2bits(j)));%⊦ (* i) i>> (* j) = (# int2bits(i) ≤ i→n(band(j, 31)) ⊃ (i < 0 ⊃ ~1;0) ;bits2int (sublist(0, # int2bits(i) - i→n(band(j, 31)), int2bits(i))) );