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Proof step 30 from proof subset-size (previous step)

¬ ¬ r ∈ st1 .refs, st1[r -- sk, val] ⊆ st0, # st0 .refs < # (st1 .refs ++ r), ∀ st. # st1 .refs ≤ # st .refs ∨ ¬ st1 ⊆ st ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ st .refs = ∅ ↔ st = ∅
Proof step 29:
# st0 .refs < # st1 .refs, st1[r -- sk, val] ⊆ st0, r ∈ st1 .refs, ∀ st. # st1 .refs ≤ # st .refs ∨ ¬ st1 ⊆ st ⊦ st1 = ∅

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ m < n → n ≠ 0], [CSIMPFORWARD : ⊦ r ∈ rset → rset ≠ ∅], [CSIMPFORWARD : ⊦ m < n → n ≠ 0], [CSIMPSEQ : ⊦ r ∈ rset → rset ++ r = rset])]]