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Proof step 10 from proof max-is-maximal (previous step)

max(ofh0) < fd, m < fd, fd ∈ ofh0[m, fh], m ≤ max(ofh0), ∀ fd0. ¬ max(ofh0) < fd0 ∨ ¬ fd0 ∈ ofh0 ⊦

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 5 terms: fd

Additional information

Used simplifier rules: ⊦ fd ∈ ofh[fd0, fh] ↔ fd = fd0 ∨ fd ∈ ofh; (ws)%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic% ⊦ m < n → ¬ n < m; (s) Used terms: fd