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Proof step 3 from proof sub-trans (previous step)

a ∈ st, a ∈ st → a ∈ st0 ∧ st[a] = st0[a], ∀ a. a ∈ st → a ∈ st0 ∧ st[a] = st0[a], ∀ a. a ∈ st0 → a ∈ st1 ∧ st0[a] = st1[a] ⊦ a ∈ st1 ∧ st[a] = st1[a]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
st[a] = st0[a], a ∈ st, a ∈ st0, ∀ a. a ∈ st → a ∈ st0 ∧ st[a] = st0[a], ∀ a. a ∈ st0 → a ∈ st1 ∧ st0[a] = st1[a] ⊦ a ∈ st1 ∧ st0[a] = st1[a]

Proof step 3 from proof sub-trans (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information