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

m0 +1 ≤ # ebatbl, ∀ ebatbl, v, evol, apebs, q. m0 ≤ # ebatbl → # absf(v, evol, apebs, ebatbl, q, m0) = # ebatbl ⊦ # absf(v, evol, apebs, ebatbl, q, m0 +1) = # ebatbl

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
m0 < # ebatbl, ∀ ebatbl, v, evol, apebs, q. m0 ≤ # ebatbl → # absf(v, evol, apebs, ebatbl, q, m0) = # ebatbl ⊦ # absf(v, evol, apebs, ebatbl, q, m0) = # ebatbl

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 ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ # evol[n, leb] = # evol; (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)