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Proof step 2 from proof abs-idset-fwd (previous step)

abs(avols, apebs0, vols, q), id(apebs1, apebs0, mapped-pebs(vols)), valid-ebatbl(apebs0, wlar, vols, n0) ⊦ abs(avols, apebs1, vols, q)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
id(apebs1, apebs0, mapped-pebs(vols)), valid-ebatbl(apebs0, wlar, vols, n0), ∀ v. (v ∈ vols ↔ v ∈ avols) ∧ ( v ∈ vols → # vols[v] = # avols[v] ∧ (∀ m. m < # vols[v] → (vols[v][m] = unmapped → ¬ avols[v][m].mapped?) ∧ ( vols[v][m] ≠ unmapped → avols[v][m].mapped? ∧ avols[v][m].data = apebs0[vols[v][m].pnum].data ∧ apebs0[vols[v][m].pnum].fill ≤ avols[v][m].fill) ∧ (avols[v][m] = erased → ¬ v × m ∈ q))) ⊦ (v ∈ vols ↔ v ∈ avols) ∧ ( v ∈ vols → # vols[v] = # avols[v] ∧ ( m < # vols[v] → (vols[v][m] = unmapped → ¬ avols[v][m].mapped?) ∧ ( vols[v][m] ≠ unmapped → avols[v][m].mapped? ∧ avols[v][m].data = apebs1[vols[v][m].pnum].data ∧ apebs1[vols[v][m].pnum].fill ≤ avols[v][m].fill) ∧ (avols[v][m] = erased → ¬ v × m ∈ q)))

Rule argument

Additional information

Used simplifier rules: abs-def: ⊦ abs(avols, apebs, vols, q) ↔ (∀ v. (v ∈ vols ↔ v ∈ avols) ∧ ( v ∈ vols → # vols[v] = # avols[v] ∧ (∀ m. m < # vols[v] → (vols[v][m] = unmapped → ¬ avols[v][m].mapped?) ∧ ( vols[v][m] ≠ unmapped → avols[v][m].mapped? ∧ avols[v][m].data = apebs[vols[v][m].pnum].data ∧ apebs[vols[v][m].pnum].fill ≤ avols[v][m].fill) ∧ (avols[v][m] = erased → ¬ v × m ∈ q))));