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Proof step 15 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)) ⊦ n0 + 1 = m, ok(nf(m)), n0 < m

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 6 terms: m

Additional information

Used simplifier rules: ls-03: ⊦ ¬ m < n → (¬ n < m ↔ n = m); from specification nat-basic1 Used terms: m