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Proof step 23 from proof to-adrs-sublist (
previous step
)
n2 + n1 ≤ # lax0, ∀ n0, n1, n. n0 + n1 ≤ # lax0 → sublist(n0, n1, lax0.to-adr(n)) = sublist(n0, n1, lax0).to-adr(n) ⊦ sublist(n2, n1, lax0.to-adr(n)) = sublist(n2, n1, lax0).to-adr(n)
Heuristic quantifier closing applied rule all left
And closed the goal!
Rule argument
pos: left 2 terms: n2, n1, n
Additional information
Used terms: n2, n1, n