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Proof step 1 from proof at-fs (conclusion)

⊦ fs[adr, nd][adr0] = (adr = adr0 ⊃ nd;fs[adr0])

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ fs[adr0, nd][adr1] = (adr0 = adr1 ⊃ nd;fs[adr1]); (s)