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Proof step 10 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)) ⊦ 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))))

User applied rule all left

And obtained the following new goals

Proof step 11:
ok(nf(n0)), ok(nf(0)), ∀ m. m < n0 → ok(nf(m)) ∧ nf(m + 1) = elemf(nf(m)), ok(nf(n0)) → rel(nf(n0), elemf(nf(n0))) ∧ ok(elemf(nf(n0))), ∀ 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

pos: left 4 terms: nf(n0) abort: F discard: F computed: T

Additional information

Used terms: nf(n0)