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Proof step 25 from proof subset-delete (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 24:
rk ∈ st, rk ∈ st → rk ∈ st0 ∧ st[rk] = st0[rk] ⊦ r ∈ st, rk ∈ st0 -- r ∧ st[rk] = (st0 -- r)[rk]

Rule argument

[EXRARG : pos: left 2, [extquanttermlist: (rk)abort:F, discard:T, computed:T]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (rk)]]