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Proof step 1 from proof sub-trans (conclusion)

⊦ st ⊆ st0 ∧ st0 ⊆ st1 → st ⊆ st1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
a ∈ st, ∀ a. a ∈ st → a ∈ st0 ∧ st[a] = st0[a], ∀ a. a ∈ st0 → a ∈ st1 ∧ st0[a] = st1[a] ⊦ a ∈ st1 ∧ st[a] = st1[a]

Rule argument

Additional information

Used simplifier rules: Subset: ⊦ st1 ⊆ st2 ↔ (∀ a. a ∈ st1 → a ∈ st2 ∧ st1[a] = st2[a]);