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Proof step 13 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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
rk ∈ st, ∀ rk. rk ∈ st → rk ∈ st0 -- r ∧ st[rk] = (st0 -- r)[rk] ⊦ r ∈ st, rk ∈ st0 ∧ st[rk] = st0[rk]

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]