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Proof step 43 from proof refs-equal (previous step)

st .refs = ∅ ∨ (∃ r. r ∈ st .refs), # st .refs = # st0 .refs, st ⊆ st0, r ∈ st0, ∀ r, st0, st1. # st1 .refs < # st .refs → (# st1 .refs = # st0 .refs ∧ r ∈ st0 ∧ st1 ⊆ st0 → r ∈ st1) ⊦ r ∈ st

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ st .refs = ∅ ↔ st = ∅
Proof step 42:
# st .refs = # st0 .refs, st ⊆ st0, r ∈ st0, st = ∅ ∨ r0 ∈ st .refs, ∀ r, st0, st1. # st1 .refs < # st .refs → (# st1 .refs = # st0 .refs ∧ r ∈ st0 ∧ st1 ⊆ st0 → r ∈ st1) ⊦ r ∈ st

Rule argument

[emptyarg]

Additional information

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