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Proof step 45 from proof subset-size (previous step)

r ∈ st0, r -- sk ∈ st0, st1 -- (r -- sk) ⊆ st0 -- (r -- sk), # st0 .refs ≤ # st1 .refs, r ∈ st0 .refs, ∀ st. # st1 .refs ≤ # st .refs ∨ ¬ st1 ⊆ st ⊦ st0 = ∅, ∀ rk. rk ∈ st1 → rk ∈ st0 ∧ st1[rk] = st0[rk], r ∈ st1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 44:
r ∈ st0 .refs, # st0 .refs ≤ # st1 .refs, st1 -- (r -- sk) ⊆ st0 -- (r -- sk), r -- sk ∈ st0, r ∈ st0, rk ∈ st1, ∀ st. # st1 .refs ≤ # st .refs ∨ ¬ st1 ⊆ st ⊦ st0 = ∅, r ∈ st1, rk ∈ st0 ∧ st1[rk] = st0[rk]

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]