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Proof step 7 from proof int2bits-falsein-02 (previous step)

¬ j < ~1, j < 0 ⊦ false ∈ int2bits(j).rest ↔ (∀ i. 2 ^ i ≠ j + 1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
∀ i. 2 ^ i ≠ 0 ⊦

Rule argument

Additional information

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