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Proof step 1 from proof sub-trans (conclusion)

⊦ stng ⊆ stng0 ∧ stng0 ⊆ stng1 → stng ⊆ stng1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
∀ a. stng[a] = stng0[a], ∀ a. stng0[a] = stng1[a] ⊦ stng[a] = stng1[a]

Rule argument

Additional information

Used simplifier rules: Subset: ⊦ stng1 ⊆ stng2 ↔ (∀ a. stng1[a] = stng2[a]);