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Proof step 2 from proof disj-in-both (previous step)

n1 < m, n0 < m, disj(asar, 0, m), adr ∈ asar[n0] ⊦ adr ∈ asar[n1] ↔ n0 = n1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
n1 < m, n0 < m, disj(asar, 0, m), adr ∈ asar[n1], adr ∈ asar[n0] ⊦ # asar[n1] = 0, # asar[n0] = 0, asar[n1] = ∅, asar[n0] = ∅, n0 = n1

Rule argument

Additional information

Used simplifier rules: size-zero-01: ⊦ as ≠ ∅ → # as ≠ 0; (forward) from specification set-basic[address-list-set]% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%ne-01: ⊦ adr ∈ as → as ≠ ∅; (forward) from specification set-basic[address-list-set]