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Proof step 9 from proof maxinf-inf (previous step)

∀ n. ∃ n0. n < n0 ∧ np(n0), max∞(np) = ⌜n1⌝ , np(n1), ∀ m. np(m) → ¬ n1 < m ⊦

User applied rule all left

And obtained the following new goals

Proof step 10:
∃ n0. n1 < n0 ∧ np(n0), max∞(np) = ⌜n1⌝ , np(n1), ∀ m. np(m) → ¬ n1 < m ⊦

Rule argument

pos: left 1 terms: n1 abort: F discard: T computed: T

Additional information

Used terms: n1