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Proof step 5 from proof uint2bits-2pot-05 (previous step)

uint2bits(i) = false ' + bits, 0 ≤ i ⊦ 2 ^ n→i(# bits) = i + 1, false ∈ bits

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
uint2bits(i) = false ' + bits, 0 ≤ i, false ∈ uint2bits(i) ⊦ 2 ^ n→i(# bits) = i + 1, false ∈ bits

Rule argument

Additional information

Used simplifier rules: uint2bits-firstpos: ⊦ uint2bits(j) = boolvar ' + bits ∧ 0 ≤ j → ¬ boolvar;% ⊦ boolvar ' + bits = bits0 → boolvar ∈ bits0; (forward)% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;