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Proof step 32 from proof refs-equal (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 31:
# st .refs = # st0 .refs, st ⊆ st0, r ∈ st0, r0 ∈ st .refs, r0 ∈ st0 .refs, ∀ r, st0, st1. # st1 .refs < # st .refs → (# st1 .refs = # st0 .refs ∧ r ∈ st0 ∧ st1 ⊆ st0 → r ∈ st1) ⊦ st = ∅, st0 = ∅, r = r0, r ∈ st, st -- r0 ⊆ st0 -- r0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]