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Proof step 15 from proof subset-size (previous step)

# st0 .refs < # st1 .refs, r ∈ st1 .refs, r ∈ st0 .refs, r ∈ st0, r -- sk ∈ st0, rk ∈ st1, ∀ st. # st1 .refs ≤ # st .refs ∨ ¬ st1 ⊆ st, rk ∈ st1 → rk ∈ st0 ∧ st1[rk] = (st0 -- (r -- sk))[rk], ∀ rk0. rk0 ∈ st1 → rk0 ∈ st0 ∧ st1[rk0] = (st0 -- (r -- sk))[rk0] ⊦ st0 = ∅, st1 = ∅, r -- sk ∈ st1, rk ∈ st0 ∧ st1[rk] = st0[rk]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
st1[rk] = (st0 -- (r -- sk))[rk], # st0 .refs < # st1 .refs, r ∈ st1 .refs, r ∈ st0 .refs, r ∈ st0, r -- sk ∈ st0, rk ∈ st1, rk ∈ st0, ∀ rk0. rk0 ∈ st1 → rk0 ∈ st0 ∧ st1[rk0] = (st0 -- (r -- sk))[rk0], ∀ st. # st1 .refs ≤ # st .refs ∨ ¬ st1 ⊆ st ⊦ st0 = ∅, st1 = ∅, st1[rk] = st0[rk], r -- sk ∈ st1

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]