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Proof step 3 from proof choice-aux (previous step)

ok(a), ∀ b. ok(b) → rel(b, elemf(b)) ∧ ok(elemf(b)) ⊦ ∃ nf. nf(0) = a ∧ ok(nf(0)) ∧ (∀ m. m < 0 → ok(nf(m)) ∧ nf(m + 1) = elemf(nf(m)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
ok(a), ∀ b. ok(b) → rel(b, elemf(b)) ∧ ok(elemf(b)) ⊦ ∃ nf. nf(0) = a

Rule argument

Additional information

Used simplifier rules: zero-01: ⊦ ¬ m < 0; from specification nat-basic1