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Proof step 7 from proof del-two (
previous step
)
a3 = a, a2 ∈ s, a1 ∈ s, a0 ∈ s, a4 ∈ s, ∀ a4. a4 = b ∨ a4 = a ∨ ¬ a4 ∈ s ⊦ a1 = a, a3 ∈ s ↔ a3 = b ∨ a3 = a, a2 = b
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 8
:
a2 ∈ s, a1 ∈ s, a0 ∈ s, a4 ∈ s, ∀ a3. a3 = b ∨ ¬ a3 ∈ s ⊦ a2 = b, a ∈ s
Rule argument
Additional information