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Proof step 8 from proof mapped-pebs-add-empty-vol (previous step)

vols[v, fill(mkebatbl(n), unmapped)][v0][m] = ⌜n0⌝ , m < # vols[v, fill(mkebatbl(n), unmapped)][v0], v0 ∈ vols[v, fill(mkebatbl(n), unmapped)] ⊦ v ∈ vols, ∃ v, m. v ∈ vols ∧ m < # vols[v] ∧ vols[v][m] = ⌜n0⌝

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 9:
vols[v, fill(mkebatbl(n), unmapped)][v0][m] = ⌜n0⌝ , m < # vols[v, fill(mkebatbl(n), unmapped)][v0], v0 ∈ vols[v, fill(mkebatbl(n), unmapped)] ⊦ v0 = v, v ∈ vols, ∃ v, m. v ∈ vols ∧ m < # vols[v] ∧ vols[v][m] = ⌜n0⌝

Rule argument

fma: v0 = v

Additional information

Used simplifier rules: ⊦ vols[v, ebatbl][v] = ebatbl; (s)% ⊦ v ∈ vols[v0, ebatbl] ↔ v = v0 ∨ v ∈ vols; (ws)% ⊦ # fill(ebatbl, ebae) = # ebatbl; (s)% ⊦ # mkebatbl(n) = n; (s)% ⊦ m < n → fill(mkebatbl(n), ebae)[m] = ebae; (s)% ⊦ unmapped ≠ ⌜m⌝ ; (s) Used formulas: v0 = v