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Proof step 2 from proof unmapped-vols-disj (previous step)

unmapped-vols(vols) ⊦ recs ⊥ vols

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
m < # vols[v], v ∈ vols, v × m ∈ recs, ∀ v, m. v ∈ vols ∧ m < # vols[v] → vols[v][m] = unmapped ⊦ vols[v][m] = unmapped

Rule argument

Additional information

Used simplifier rules: unmapped-vols: ⊦ unmapped-vols(vols) ↔ (∀ v, m. v ∈ vols ∧ m < # vols[v] → vols[v][m] = unmapped);%disj: ⊦ recs ⊥ vols ↔ (∀ v, m. ¬ v × m ∈ recs ∨ ¬ v ∈ vols ∨ ¬ m < # vols[v] ∨ vols[v][m] = unmapped);% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic