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Proof step 9 from proof max-is-maximal (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
max(ofh0) < fd, m < fd, fd ∈ ofh0[m, fh], m ≤ max(ofh0), ∀ fd0. ¬ max(ofh0) < fd0 ∨ ¬ fd0 ∈ ofh0 ⊦

Rule argument

Additional information

Used simplifier rules: fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat% ⊦ m < n ↔ ¬ n ≤ m; (ss)