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Proof step 6 from proof uint2bits-onein (previous step)

uint2bits(j) = boolvar + bits ∧ uint2bits(j) ≠ [], 0 < j, bits.first ⊦ true ∈ uint2bits(j)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
uint2bits(j) = boolvar ' + bits, 0 < j, bits.first, boolvar ∈ uint2bits(j) ⊦ true ∈ uint2bits(j)

Rule argument

Additional information

Used simplifier rules: uint2bits-empty: 0 ≤ i ⊦ uint2bits(i) ≠ [];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ' + bits = bits0 → boolvar ∈ bits0; (forward)% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i < j → ¬ j < i;% ⊦ boolvar + bits = boolvar ' + bits;