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Proof step 4 from proof subset-put-notin (previous step)

⊦ st ⊆ st0 ∧ ¬ r -- sk ∈ st ∨ r -- sk ∈ st ∧ st[r -- sk] = val ∧ st -- (r -- sk) ⊆ st0 -- (r -- sk) ↔ st ⊆ st0, r ∈ st

User applied rule case distinction

And obtained the following new goals

Proof step 2:
⊦ r -- sk ∈ st → r ∈ st
Proof step 3:
⊦ r -- sk ∈ st → r ∈ st

Rule argument

[CASEDARG : pos: right 1, ()]

Additional information

[[CONCRETESIMPRULES : ()]]