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Proof step 19 from proof choice-aux (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
ok(nf(n0)), ok(nf(0)), rel(nf(n0), elemf(nf(n0))), ok(elemf(nf(n0))), m < n0, ∀ b. ok(b) → rel(b, elemf(b)) ∧ ok(elemf(b)), ∀ m. m < n0 → ok(nf(m)) ∧ nf(m + 1) = elemf(nf(m)) ⊦ nf(m + 1) = elemf(nf(m)), n0 = m, n0 = 0, n0 + 1 = m

Proof step 19 from proof choice-aux (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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