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Proof step 12 from proof fids-cons-reduce (previous step)

∀ fid. (fid ∈ fs0 ↔ fid ∈ tree0.fids ∨ fid ∈ ofh0.fids) → (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 all left

And obtained the following new goals

Proof step 13:
(fid ∈ fs0 ↔ fid ∈ tree0.fids ∨ fid ∈ ofh0.fids) → (fid ∈ fs1 ↔ fid ∈ tree1.fids ∨ fid ∈ ofh1.fids), ∀ 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 ∈ fs1 ↔ fid ∈ tree1.fids ∨ fid ∈ ofh1.fids

Rule argument

pos: left 1 terms: fid abort: F discard: F computed: T

Additional information

Used terms: fid