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

⊦ (i = 32768 ∧ j = 1 → i2s(i i>> j) ≠ i2s(s2i(i2s(i)) i>> j)) ∧ (i = 32768 ∧ j = 1 → i2s(i i>>> j) ≠ i2s(s2i(i2s(i)) i>>> j))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ ( i = 32768 ∧ j = 1 → i2s ((# int2bits(32768) ≤ i→n(bits2int(bitsand(int2bits(1), int2bits(31)))) ⊃ 0 ;bits2int(sublist(0, # int2bits(32768) - i→n(bits2int(bitsand(int2bits(1), int2bits(31)))), int2bits(32768))) )) ≠ i2s ((# int2bits(~32768) ≤ i→n(bits2int(bitsand(int2bits(1), int2bits(31)))) ⊃ ~1 ;bits2int(sublist(0, # int2bits(~32768) - i→n(bits2int(bitsand(int2bits(1), int2bits(31)))), int2bits(~32768))) ))) ∧ (i = 32768 ∧ j = 1 → i2s(32768 i>>> 1) ≠ i2s(~32768 i>>> 1))

Rule argument

Additional information

Used simplifier rules: sis-pos-high: ⊦ 0 ≤ i ∧ 32767 < i % 65536 → s2i(i2s(i)) = i % 65536 - 65536;% ⊦ 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))) );