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Proof step 13 from proof short-law-righthift-not (previous step)

0 ≤ 31 ⊦ ( i = 32768 ∧ j = 1 → i2s ((# int2bits(32768) ≤ i→n(bits2int(bitsand(false ' + true ', uint2bits(31)))) ⊃ 0 ;bits2int(sublist(0, # int2bits(32768) - i→n(bits2int(bitsand(false ' + true ', uint2bits(31)))), int2bits(32768))) )) ≠ i2s ((# int2bits(~32768) ≤ i→n(bits2int(bitsand(false ' + true ', uint2bits(31)))) ⊃ ~1 ;bits2int(sublist(0, # int2bits(~32768) - i→n(bits2int(bitsand(false ' + true ', uint2bits(31)))), int2bits(~32768))) ))) ∧ ( i = 32768 ∧ j = 1 → i2s (bits2int (sublist (0, # fillsgn(32, int2bits(32768)), mklist(false, i→n(bits2int(bitsand(false ' + true ', uint2bits(31))))) + fillsgn(32, int2bits(32768))))) ≠ i2s (bits2int (sublist (0, # fillsgn(32, int2bits(~32768)), mklist(false, i→n(bits2int(bitsand(false ' + true ', uint2bits(31))))) + fillsgn(32, int2bits(~32768))))))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ bits + [] = bits;% ⊦ mklist(boolvar, 0) = [];%n ≠ 0 ⊦ mklist(false, (* n)) = false ' + mklist(false, n -1);% ⊦ bits2uint(true ') = 1;% ⊦ bits2int(false ' + bits) = bits2uint(bits);% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bitsand(false ', bits + boolvar ') = false ';%⊦ bitsand(boolvar ' + boolvar0 ', bits0 + boolvar1 ' + boolvar2 ') = bitsand(boolvar ', bits0 + boolvar1 ') + (boolvar0 ∧ boolvar2);% ⊦ bits + boolvar = bits + boolvar ';% ⊦ boolvar + boolvar0 = boolvar ' + boolvar0 ';% ⊦ uint2bits(1) = false + true;%1 < i ⊦ uint2bits((* i)) = uint2bits(i / 2) + (i % 2 = 1);