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Proof step 2 from proof disj-remove (previous step)

v × m ∈ recs, recs ⊥ vols ⊦ recs -- (v × m) ⊥ vols

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
m0 < # vols[v0], v0 ∈ vols, v0 × m0 ∈ recs, v × m ∈ recs, ∀ v, m. ¬ v × m ∈ recs ∨ ¬ v ∈ vols ∨ ¬ m < # vols[v] ∨ vols[v][m] = unmapped ⊦ vols[v0][m0] = unmapped, v0 × m0 = v × m

Rule argument

Additional information

Used simplifier rules: 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% ⊦ ladr ∈ recs -- ladr0 ↔ ladr ≠ ladr0 ∧ ladr ∈ recs; (s)