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Proof step 6 from proof Extension (previous step)

∀ a. (a ∈ st1 ↔ a ∈ st2) ∧ lookup(a, st1) = lookup(a, st2) ⊦ ∀ a. (a ∈ st1 ↔ a ∈ st2) ∧ st1[a] = st2[a]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
∀ a. (a ∈ st1 ↔ a ∈ st2) ∧ lookup(a, st1) = lookup(a, st2) ⊦ (a ∈ st1 ↔ a ∈ st2) ∧ lookup(a, st1) = lookup(a, st2)

Rule argument

Additional information

Used simplifier rules: lookup-def: ⊦ st[a] = lookup(a, st); from specification lookup