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Proof step 43 from proof subset-size (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 42:
r ∈ st0 .refs, # st0 .refs ≤ # st1 .refs, r -- sk ∈ st0, r ∈ st0, rk ∈ st1, ∀ st. # st1 .refs ≤ # st .refs ∨ ¬ st1 ⊆ st, ∀ rk0. rk0 ∈ st1 -- (r -- sk) → rk0 ≠ r -- sk ∧ rk0 ∈ st0 ∧ st1[rk0] = st0[rk0] ⊦ st0 = ∅, r ∈ st1, rk ∈ st0 ∧ st1[rk] = st0[rk]

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ rk ≠ rk0 → (st -- rk0)[rk] = st[rk]], [CSIMPSEQ : ⊦ rk ∈ st -- rk0 ↔ rk ≠ rk0 ∧ rk ∈ st])]]