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Proof step 6 from proof subset-notin (previous step)

∃ sk. r -- sk ∈ st, ∀ rk. rk ∈ st → rk ∈ st0 ∧ st[rk] = st0[rk] ⊦ r ∈ st0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⊦ r -- sk ∈ st → r ∈ st
Proof step 5:
r -- sk ∈ st, r ∈ st, ∀ rk. rk ∈ st → rk ∈ st0 ∧ st[rk] = st0[rk] ⊦ r ∈ st0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]