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Proof step 5 from proof in-write (previous step)

# ax0 = # ndx0, ∀ ndx, adr, fs. # ndx0 = # ndx → (adr ∈ fs[ax0, ndx] ↔ adr ∈ fs ∨ adr ∈ ax0) ⊦ adr ∈ fs[adr0, nd][ax0, ndx0] ↔ adr ∈ fs ∨ adr = adr0 ∨ adr ∈ ax0

User applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: ndx0, adr, fs[adr0, nd] abort: F discard: T computed: T

Additional information

Used simplifier rules: ⊦ adr ∈ fs[adr1, nd] ↔ adr = adr1 ∨ adr ∈ fs; (s) Used terms: ndx0, adr, fs[adr0, nd]