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Proof step 6 from proof max-is-maximal (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 7:
m < fd, max(ofh0) < m, fd ∈ ofh0[m, fh], ∀ fd0. ¬ max(ofh0) < fd0 ∨ ¬ fd0 ∈ ofh0 ⊦
Proof step 9:
max(ofh0) < fd, fd ∈ ofh0[m, fh], ∀ fd0. ¬ max(ofh0) < fd0 ∨ ¬ fd0 ∈ ofh0 ⊦ max(ofh0) < m

Rule argument

pos: left 1

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