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Proof step 26 from proof refs-equal (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 25:
# st .refs = # st0 .refs, r0 ∈ st0 .refs, r0 ∈ st .refs, r ∈ st0, rk ∈ st -- r0, rk ∈ st → rk ∈ st0 ∧ st[rk] = st0[rk], ∀ rk. rk ∈ st → rk ∈ st0 ∧ st[rk] = st0[rk], ∀ r, st0, st1. # st1 .refs < # st .refs → (# st1 .refs = # st0 .refs ∧ r ∈ st0 ∧ st1 ⊆ st0 → r ∈ st1) ⊦ r = r0, st0 = ∅, st = ∅, r0 = rk .ref, r ∈ st, rk ∈ st0 -- r0 ∧ st[rk] = st0[rk]

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ r ≠ rk .ref → (st -- r)[rk] = st[rk]])]]