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Proof step 4 from proof Extension (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
∀ a0. (a0 ∈ st1 ↔ a0 ∈ st2) ∧ lookup(a0, st1) = lookup(a0, 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