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Proof step 78 from proof subset-is-union (conclusion)

⊦ st ⊆ st1 ↔ (∃ st0. st1 = st0 ∪ st ∧ st .keys ∩ st0 .keys = ∅)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 1:
⊦ st0 ⊆ st ∪ st0
Proof step 77:
st ⊆ st1 ⊦ ∃ st0. st1 = st0 ∪ st ∧ st .keys ∩ st0 .keys = ∅

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]