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Proof step 36 from proof refs-equal (previous step)

# st .refs = # st0 .refs, r0 ∈ st .refs, r ∈ st0, st ⊆ st0, ∀ r, st0, st1. # st1 .refs < # st .refs → (# st1 .refs = # st0 .refs ∧ r ∈ st0 ∧ st1 ⊆ st0 → r ∈ st1) ⊦ # (st -- r0) .refs = # (st0 -- r0) .refs ∧ r ∈ st0 -- r0 ∧ st -- r0 ⊆ st0 -- r0, st = ∅, r ∈ st

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
⊦ (st -- r) .refs = st .refs -- r
Proof step 35:
# st .refs = # st0 .refs, r0 ∈ st .refs, r ∈ st0, st ⊆ st0, ∀ r, st0, st1. # st1 .refs < # st .refs → (# st1 .refs = # st0 .refs ∧ r ∈ st0 ∧ st1 ⊆ st0 → r ∈ st1) ⊦ st = ∅, r ∈ st, # st0 .refs - 1 = # (st0 .refs -- r0) ∧ r ∈ st0 -- r0 ∧ st -- r0 ⊆ st0 -- r0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ r ∈ rset → # (rset -- r) = # rset - 1])]]