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Proof step 12 from proof absf-at (previous step)

n < m0, m0 < # ebatbl, n < # ebatbl, n + 1 < # ebatbl, ∀ n, ebatbl, v, evol, apebs, q. n < m0 ∧ n < # ebatbl ∧ m0 ≤ # ebatbl → absf(v, evol, apebs, ebatbl, q, m0)[n] = (ebatbl[n] = unmapped ⊃ (v × n ∈ q ⊃ unmapped;erased) ;mapped(apebs[ebatbl[n].pnum].data, (evol[n].mapped? ⊃ evol[n].fill;apebs[ebatbl[n].pnum].fill)) ) ⊦ absf(v, evol, apebs, ebatbl, q, m0)[n] = (ebatbl[n] = unmapped ⊃ (v × n ∈ q ⊃ unmapped;erased) ;mapped(apebs[ebatbl[n].pnum].data, (evol[n].mapped? ⊃ evol[n].fill;apebs[ebatbl[n].pnum].fill)) ), n = m0

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 5 terms: n, ebatbl, v, evol, apebs, q

Additional information

Used simplifier rules: ⊦ m < n → m ≤ n; (s) Used terms: n, ebatbl, v, evol, apebs, q