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Proof step 5 from proof choice-aux (previous step)

∀ a, elemf. ok(a) ∧ (∀ b. ok(b) → rel(b, elemf(b)) ∧ ok(elemf(b))) → (∃ nf. nf(0) = a ∧ ok(nf(n0)) ∧ (∀ m. m < n0 → ok(nf(m)) ∧ nf(m + 1) = elemf(nf(m)))), ok(a), ∀ b. ok(b) → rel(b, elemf(b)) ∧ ok(elemf(b)) ⊦ ∃ nf. nf(0) = a ∧ ok(nf(n0 +1)) ∧ (∀ m. m < n0 +1 → ok(nf(m)) ∧ nf(m + 1) = elemf(nf(m)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
ok(a), ∀ a, elemf. ok(a) ∧ (∀ b. ok(b) → rel(b, elemf(b)) ∧ ok(elemf(b))) → (∃ nf. nf(0) = a ∧ ok(nf(n0)) ∧ (∀ m. m < n0 → ok(nf(m)) ∧ nf(m + 1) = elemf(nf(m)))), ∀ b. ok(b) → rel(b, elemf(b)) ∧ ok(elemf(b)) ⊦ ∃ nf. nf(0) = a ∧ ok(nf(n0 + 1)) ∧ (∀ m. ¬ n0 < m → ok(nf(m)) ∧ nf(m + 1) = elemf(nf(m)))

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%succ: ⊦ m +1 = m + 1; from specification nat-basic2