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Proof step 9 from proof uint2bits-expos0 (previous step)

uint2bits(i) = false ' + bits1, false ∈ uint2bits(i), 0 < i ⊦ uint2bits(i) = false ' + true ' + bits1.rest, ∃ bits. uint2bits(i) = false ' + true ' + bits

User applied rule insert lemma

And obtained the following new goals

Proof step 10:
⊦ 0 < i → uint2bits(i).rest.first
Proof step 11:
0 < i → uint2bits(i).rest.first, uint2bits(i) = false ' + bits1, false ∈ uint2bits(i), 0 < i ⊦ uint2bits(i) = false ' + true ' + bits1.rest, ∃ bits. uint2bits(i) = false ' + true ' + bits

Rule argument

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

Additional information

Used terms: i