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Proof step 1 from proof uint2bits-expos1 (conclusion)

1 < j ⊦ ∃ bits. uint2bits(j) = false ' + true ' + bits ∧ bits ≠ []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
1 < j, 0 < j, 0 ≤ j ⊦ ∃ bits. uint2bits(j) = false ' + true ' + bits ∧ bits ≠ []

Rule argument

Additional information

Used simplifier rules: ⊦ (* i) < j ∧ 0 < i → 0 ≤ j ∧ 0 < j; (forward)