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Proof step 13 from proof maxinf-fin (previous step)

max∞(np) = ⌜n0⌝ , np(n0), ∀ m. np(m) → ¬ max∞(np) < ⌜m⌝ ⊦ max∞(np) = ⌜n⌝ ↔ np(n) ∧ (∀ m. np(m) → ¬ n < m), max∞(np) = ∞

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
max∞(np) = ⌜n0⌝ , np(n0), ∀ m. np(m) → ¬ n0 < m ⊦ n0 = n ↔ np(n) ∧ (∀ m. np(m) → ¬ n < m)

Rule argument

Additional information

Used simplifier rules: less-fin: ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; from specification nat-inf%inj: ⊦ ⌜m⌝ = ⌜m0⌝ ↔ m = m0; from specification nat-inf-basic%disj: ⊦ ⌜m⌝ ≠ ∞; from specification nat-inf-basic