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Proof step 6 from proof f (
previous step
)
dups(x0 + a '), dups(x0 + a0 ') → a0 ∈ x0, ∀ a1. dups(x0 + a1 ') → a1 ∈ x0 ⊦ a = a0, dups(x0), a0 ∈ x0, a ∈ x0
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 7
:
dups(x0 + a '), ∀ a1. dups(x0 + a1 ') → a1 ∈ x0 ⊦ dups(x0), a0 ∈ x0, a ∈ x0, dups(x0 + a0 ')
Rule argument
Additional information