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Proof step 13 from proof choice-aux (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 14:
n0 + 1 ≠ m, ok(nf(n0)), ok(nf(0)), rel(nf(n0), elemf(nf(n0))), ok(elemf(nf(n0))), ∀ b. ok(b) → rel(b, elemf(b)) ∧ ok(elemf(b)), ∀ m. m < n0 → ok(nf(m)) ∧ nf(m + 1) = elemf(nf(m)) ⊦ ok(nf(m)), n0 < m

Rule argument

pos: right 1

Proof step 13 from proof choice-aux (previous step)

User applied rule case distinction

And obtained the following new goals

Rule argument

Additional information