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Proof step 1 from proof int2bits-expos (conclusion)

0 < i ⊦ ∃ bits. int2bits(i) = false ' + true ' + bits

User applied rule insert lemma

And obtained the following new goals

Proof step 2:
0 < i ⊦ ∃ bits. uint2bits(i) = false ' + true ' + bits
Proof step 3:
∃ bits. uint2bits(i) = false ' + true ' + bits, 0 < i ⊦ ∃ bits. int2bits(i) = false ' + true ' + bits

Rule argument

name: uint2bits-expos0 subst: [i] → [i]

Additional information

Used terms: i