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Proof step 46 from proof refs-equal (previous step)

# st .refs = # st0 .refs, r ∈ st0, st ⊆ st0 ⊦ r ∈ st

User applied rule induction

And obtained the following new goals

Proof step 45:
# st .refs = # st0 .refs ∧ r ∈ st0 ∧ st ⊆ st0, ∀ r, st0, st1. # st1 .refs < # st .refs → (# st1 .refs = # st0 .refs ∧ r ∈ st0 ∧ st1 ⊆ st0 → r ∈ st1) ⊦ r ∈ st

Rule argument

[INDHYPARG : [STANDARDINDUCTIONTYPE : ], true, false, # st .refs, [] → [], <]

Additional information

[[CONCRETESIMPRULES : ([CSIMPLESSPRED : <])], [EXTRATERMS : (# st .refs)]]