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Proof step 2 from proof absf-at (previous step)

n < m, m ≤ # ebatbl ⊦ absf(v, evol, apebs, ebatbl, q, m)[n] = (ebatbl[n] = unmapped ⊃ (v × n ∈ q.lebrefs ⊃ unmapped;erased) ;mapped(apebs[ebatbl[n].pnum].data, (evol[n].mapped? ⊃ evol[n].fill;apebs[ebatbl[n].pnum].fill)) )

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
n < m, n < # ebatbl, m ≤ # ebatbl ⊦ absf(v, evol, apebs, ebatbl, q, m)[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)) )

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ none ≠ v × m; (s)% ⊦ lref ∈ q0.lebrefs ↔ lref ∈ q0 ∧ lref ≠ none; (s)