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Proof step 8 from proof choice-aux (previous step)

ok(nf(n0)), ok(nf(0)), ∀ m. m < n0 → ok(nf(m)) ∧ nf(m + 1) = elemf(nf(m)), ∀ b. ok(b) → rel(b, elemf(b)) ∧ ok(elemf(b)) ⊦ ∃ nf0. nf0(0) = nf(0) ∧ ok(nf0(n0 + 1)) ∧ (∀ m. ¬ n0 < m → ok(nf0(m)) ∧ nf0(m + 1) = elemf(nf0(m)))

User applied rule exists right

And obtained the following new goals

Proof step 9:
ok(nf(n0)), ok(nf(0)), ∀ m. m < n0 → ok(nf(m)) ∧ nf(m + 1) = elemf(nf(m)), ∀ b. ok(b) → rel(b, elemf(b)) ∧ ok(elemf(b)) ⊦ nf(n0 + 1; elemf(nf(n0)))(0) = nf(0) ∧ ok(nf(n0 + 1; elemf(nf(n0)))(n0 + 1)) ∧ (∀ m. ¬ n0 < m → ok(nf(n0 + 1; elemf(nf(n0)))(m)) ∧ nf(n0 + 1; elemf(nf(n0)))(m + 1) = elemf(nf(n0 + 1; elemf(nf(n0)))(m)))

Rule argument

pos: right 1 terms: nf(n0 + 1; elemf(nf(n0))) abort: F discard: T computed: F

Additional information

Used terms: nf(n0 + 1; elemf(nf(n0)))