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Proof step 6 from proof choice-aux (previous step)

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)))

User applied rule all left

And obtained the following new goals

Proof step 7:
ok(a), 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

pos: left 2 terms: a, elemf abort: F discard: T computed: T

Additional information

Used terms: a, elemf