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Proof step 18 from proof choice-aux (previous step)

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((n0 + 1 = m ⊃ elemf(nf(n0));nf(m))), n0 = m, n0 = 0

User applied rule case distinction

And obtained the following new goals

Proof step 19:
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

Rule argument

pos: right 1

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%eq: ⊦ m = m + n ↔ n = 0; from specification nat-basic2%l-02: ⊦ ¬ m + n < m; from specification nat-basic2