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Proof step 4 from proof k-Extension (
previous step
)
∀ a. (a ∈ st1 ↔ a ∈ st2) ∧ (a ∈ st1 → st1[a] = st2[a]) ⊦ ∀ a. (a ∈ st1 ↔ a ∈ st2) ∧ st1[a] = st2[a]
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 5
:
∀ a. (a ∈ st1 ↔ a ∈ st2) ∧ (a ∈ st1 → st1[a] = st2[a]) ⊦ (a ∈ st1 ↔ a ∈ st2) ∧ st1[a] = st2[a]
Rule argument
Additional information