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Proof step 5 from proof read-write-other-fs (previous step)

∀ fs, adr, nd. fs[adr, nd][ax0] = fs[ax0] ∨ adr ∈ ax0 ⊦ fs[adr, nd][ax0] = fs[ax0], adr = adr0, adr ∈ ax0

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 1 terms: fs, adr, nd

Additional information

Used terms: fs, adr, nd