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Proof step 9 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)) ⊦ 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)))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
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)) ⊦ ok(elemf(nf(n0))) ∧ (¬ n0 < m → ok((n0 + 1 = m ⊃ elemf(nf(n0));nf(m))) ∧ (n0 = m ⊃ elemf(nf(m));nf(m + 1)) = elemf((n0 + 1 = m ⊃ elemf(nf(n0));nf(m))))

Rule argument

Additional information

Used simplifier rules: eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2