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Proof step 4 from proof absf-at (previous step)

n < # ebatbl, n < m0 +1, m0 +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 +1)[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)) )

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
n < # ebatbl, m0 < # ebatbl, n ≤ m0, ∀ 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) [m0, (ebatbl[m0] = unmapped ⊃ (v × m0 ∈ q ⊃ unmapped;erased) ;mapped(apebs[ebatbl[m0].pnum].data, (evol[m0].mapped? ⊃ evol[m0].fill;apebs[ebatbl[m0].pnum].fill)) ) ] [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: absf-rec: ⊦ n < # ebatbl → absf(v, evol, apebs, ebatbl, q, n + 1) = absf(v, evol, apebs, ebatbl, q, n) [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)) ) ];% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ none ≠ v × m; (s)% ⊦ lref ∈ q0.lebrefs ↔ lref ∈ q0 ∧ lref ≠ none; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≤ n ↔ m < n; (s)% ⊦ m +1 = m + 1; (s)