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Proof step 4 from proof uint2bits-expos0 (previous step)

uint2bits(i) = boolvar ' + bits0, 0 < i, boolvar ∈ uint2bits(i) ⊦ uint2bits(i) = false ' + true ' + bits0.rest, ∃ bits. uint2bits(i) = false ' + true ' + bits

User applied rule insert lemma

And obtained the following new goals

Proof step 5:
⊦ 0 ≤ i → ¬ uint2bits(i).first
Proof step 6:
0 ≤ i → ¬ uint2bits(i).first, uint2bits(i) = boolvar ' + bits0, 0 < i, boolvar ∈ uint2bits(i) ⊦ uint2bits(i) = false ' + true ' + bits0.rest, ∃ bits. uint2bits(i) = false ' + true ' + bits

Rule argument

name: uint2bits-first subst: [i] → [i]

Additional information

Used terms: i