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Proof step 15 from proof to-adrs-sublist (previous step)

m1 ≤ # lax0, ∀ n0, n1, n. n0 + n1 ≤ # lax0 → sublist(n0, n1, lax0.to-adr(n)) = sublist(n0, n1, lax0).to-adr(n) ⊦ sublist(0, m1, lax0.to-adr(n)) = sublist(0, m1, lax0).to-adr(n)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: 0, m1, n

Additional information

Used simplifier rules: ⊦ n + 0 = n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used terms: 0, m1, n