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Proof step 13 from proof union-subset-left (previous step)

st .refs ∩ st0 .refs = ∅ ⊦ ∀ rk. rk ∈ st → rk ∈ st ∪ st0 ∧ st[rk] = (st ∪ st0)[rk]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ rk ∈ st ∪ st0 ↔ rk ∈ st ∨ rk ∈ st0
Proof step 12:
st .refs ∩ st0 .refs = ∅, rk ∈ st ⊦ st[rk] = (st ∪ st0)[rk]

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]