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Proof step 8 from proof fids-cons-reduce (previous step)

∀ fid. fid ∈ fs0 ↔ fid ∈ tree0.fids ∨ fid ∈ ofh0.fids ⊦ (∀ fid. fid ∈ fs1 ↔ fid ∈ tree1.fids ∨ fid ∈ ofh1.fids) ↔ (∀ fid. (fid ∈ fs0 ↔ fid ∈ tree0.fids ∨ fid ∈ ofh0.fids) → (fid ∈ fs1 ↔ fid ∈ tree1.fids ∨ fid ∈ ofh1.fids))

User applied rule case distinction

And obtained the following new goals

Proof step 9:
∀ fid. fid ∈ fs0 ↔ fid ∈ tree0.fids ∨ fid ∈ ofh0.fids, ∀ fid. fid ∈ fs1 ↔ fid ∈ tree1.fids ∨ fid ∈ ofh1.fids ⊦ ∀ fid. (fid ∈ fs0 ↔ fid ∈ tree0.fids ∨ fid ∈ ofh0.fids) → (fid ∈ fs1 ↔ fid ∈ tree1.fids ∨ fid ∈ ofh1.fids)
Proof step 11:
∀ fid. fid ∈ fs0 ↔ fid ∈ tree0.fids ∨ fid ∈ ofh0.fids, ∀ fid. (fid ∈ fs0 ↔ fid ∈ tree0.fids ∨ fid ∈ ofh0.fids) → (fid ∈ fs1 ↔ fid ∈ tree1.fids ∨ fid ∈ ofh1.fids) ⊦ ∀ fid. fid ∈ fs1 ↔ fid ∈ tree1.fids ∨ fid ∈ ofh1.fids

Rule argument

pos: right 1

Proof step 8 from proof fids-cons-reduce (previous step)

User applied rule case distinction

And obtained the following new goals

Rule argument

Additional information