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Proof step 6 from proof ls-valid-global-sqn-vtbl (previous step)

valid-vtbl(apebs, vols, wlar, vtblpnum), valid-wl(apebs, wlar), sqnum ≤ valid-sqns(VTBL_VOLID, VTBL_LNUM, apebs).max, ∀ v, m. valid-sqns(v, m, apebs) ≠ ∅ → valid-sqns(v, m, apebs).max < sqnum ⊦

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: VTBL_VOLID, VTBL_LNUM

Additional information

Used simplifier rules: ⊦ valid-vtbl(apebs, vols, wlar, n0) ∧ valid-wl(apebs, wlar) → valid-sqns(VTBL_VOLID, VTBL_LNUM, apebs) ≠ ∅; (s)% ⊦ m ≤ n → (n < m ↔ false); (s) Used terms: VTBL_VOLID, VTBL_LNUM