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Proof step 42 from proof refs-equal (previous step)

# 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

User applied rule cut formula

And obtained the following new goals

Proof step 3:
⊦ ∅ .refs = ∅
Proof step 4:
⊦ st .refs = ∅ ↔ st = ∅
Proof step 41:
# 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) ⊦ st = ∅, r ∈ st

Rule argument

[fma-arg: st = ∅]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ # ∅ = 0], [CSIMPSEQ : ⊦ ∅ ⊆ st], [CSIMPSEQ : ⊦ ¬ r ∈ ∅], [CSIMPFORWARD : ⊦ # rset = 0 → rset = ∅])], [EXTRAFMAS : (st = ∅)]]