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Proof step 1 from proof int2bits-rest (conclusion)

⊦ int2bits(j).rest = [] ↔ j = 0 ∨ j = ~1

User applied rule case distinction

And obtained the following new goals

Proof step 2:
int2bits(j).rest = [] ⊦ j = 0 ∨ j = ~1

Rule argument

pos: right 1

Additional information

Used simplifier rules: int2bits-zero: ⊦ int2bits(0) = false ';%int2bits-neg-num: i < 0 ⊦ int2bits((* i)) = sgn(bitsinvert(uint2bits(abs(i))) +1);%uint2bits-one: ⊦ uint2bits(1) = false + true;% ⊦ boolvar '.rest = [];% ⊦ boolvar + boolvar0 = boolvar ' + boolvar0 ';% ⊦ bitsinvert(false ' + bits) = true + bitsinvert(bits);% ⊦ bitsinvert(true ') = false ';% ⊦ boolvar + bits = boolvar ' + bits;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ (bits + false ')+1 = bits + true;% ⊦ bits + boolvar = bits + boolvar ';% ⊦ sgn(true ' + true ') = true ';