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Proof step 3 from proof max-is-maximal (previous step)

max(ofh) < fd, fd ∈ ofh ⊦

User applied rule structural induction

And obtained the following new goals

Proof step 4:
max(ofh0[m, fh]) < fd, fd ∈ ofh0[m, fh], ∀ fd. ¬(max(ofh0) < fd ∧ fd ∈ ofh0) ⊦

Rule argument

indvar: ofh

Additional information

Used simplifier rules: open-file-handles generated by ∅, [ ];;% ⊦ ¬ fd ∈ ∅; (s)