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Proof step 27 from proof subset-delete (previous step)

⊦ (∀ rk. rk ∈ st → rk ∈ st0 -- r ∧ st[rk] = (st0 -- r)[rk]) ↔ (∀ rk. rk ∈ st → rk ∈ st0 ∧ st[rk] = st0[rk]), r ∈ st

User applied rule case distinction

And obtained the following new goals

Proof step 13:
∀ rk. rk ∈ st → rk ∈ st0 -- r ∧ st[rk] = (st0 -- r)[rk] ⊦ ∀ rk. rk ∈ st → rk ∈ st0 ∧ st[rk] = st0[rk], r ∈ st
Proof step 26:
∀ rk. rk ∈ st → rk ∈ st0 ∧ st[rk] = st0[rk] ⊦ ∀ rk. rk ∈ st → rk ∈ st0 -- r ∧ st[rk] = (st0 -- r)[rk], r ∈ st

Rule argument

[CASEDARG : pos: right 1, ()]

Additional information

[[CONCRETESIMPRULES : ()]]