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Proof step 4 from proof refcount-addobj (conclusion)

⊦ # addobj(newref(st), class, fis, st) .refs = # st .refs + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 1:
⊦ addobj(r, class, fis, st) .refs = st .refs ++ r
Proof step 3:
newref(st) ∈ st .refs ⊦ st = ∅

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ st .refs = ∅ ↔ st = ∅], [CSIMPFORWARD : ⊦ r ∈ rset → rset ≠ ∅], [CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ # rset + 1 = # (rset ++ r) ↔ ¬ r ∈ rset])]]