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

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

User applied rule case distinction

And obtained the following new goals

Proof step 19:
m < n0, ok(elemf(nf(n0))), ok(nf(0)), ok(nf(n0)), rel(nf(n0), 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, n0 = m, nf(m + 1) = elemf(nf(m))

Rule argument

pos: right 2

Additional information

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