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Proof step 7 from proof size-two-06 (
previous step
)
#(s0 ++ a0) < 2, a ∈ s0, b = a0 ∨ b ∈ s0, ∀ b, a. # s0 < 2 ∧ b ∈ s0 ∧ a ∈ s0 → a = b ⊦ b = a, a = b
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 8
:
#(s0 ++ a0) < 2, a ∈ s0, b = a0 ∨ b ∈ s0, ∀ b, a. # s0 < 2 ∧ b ∈ s0 ∧ a ∈ s0 → a = b ⊦ a = b
Rule argument
Additional information