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Proof step 28 from proof subset-size (previous step)

# st0 .refs < # st1 .refs, st1 -- (r -- sk) ⊆ st0 -- (r -- sk) ∧ (r -- sk) × val ∈ st0, r ∈ st1 .refs, ∀ st. # st1 .refs ≤ # st .refs ∨ ¬ st1 ⊆ st ⊦ st1 = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 27:
r ∈ st0, r -- sk ∈ st0, st1 -- (r -- sk) ⊆ st0 -- (r -- sk), # st0 .refs < # st1 .refs, r ∈ st1 .refs, ∀ st. # st1 .refs ≤ # st .refs ∨ ¬ st1 ⊆ st ⊦ st1 = ∅

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPFORWARD : ⊦ r -- sk ∈ st → r ∈ st], [CSIMPSEQ : ⊦ rk × val ∈ st ↔ rk ∈ st ∧ st[rk] = val])]]