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Proof step 20 from proof newrefs-three (previous step)

¬ newref(st[newref(st) -- 0 ', noval][newref(st[newref(st) -- 0 ', noval]) -- 0 ', noval]) -- sk ≠ newref(st[newref(st) -- 0 ', noval]) -- 0 ', ¬ newref(st[newref(st) -- 0 ', noval][newref(st[newref(st) -- 0 ', noval]) -- 0 ', noval]) -- sk ∈ (st[newref(st) -- 0 ', noval][newref(st[newref(st) -- 0 ', noval]) -- 0 ', noval]) ⊦ ¬ newref(st[newref(st) -- 0 ', noval][newref(st[newref(st) -- 0 ', noval]) -- 0 ', noval]) -- sk ∈ st

User applied rule simplifier

And obtained the following new goals

Proof step 15:
⊦ ¬ newref(st) -- sk ∈ st
Proof step 16:
⊦ r -- sk ∈ st → r ∈ st
Proof step 19:
newref(st[newref(st) -- 0 ', noval][newref(st[newref(st) -- 0 ', noval]) -- 0 ', noval]) = newref(st[newref(st) -- 0 ', noval]), newref(st[newref(st) -- 0 ', noval]) ∈ st, newref(st[newref(st) -- 0 ', noval]) -- 0 ' ∈ st ⊦

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ r -- sk = r0 -- sk0 ↔ r = r0 ∧ sk = sk0])]]