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Proof step 5 from proof Diff-union-comm (previous step)

key ∈ (ri \ ks0) \ ks1 ↔ key ∈ ri \ (ks0 ∪ ks1) ⊦ (key ∈ (ri \ ks0) \ ks1 ↔ key ∈ ri \ (ks0 ∪ ks1)) ∧ (key ∈ (ri \ ks0) \ ks1 → ((ri \ ks0) \ ks1)[key] = (ri \ (ks0 ∪ ks1))[key])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
key ∈ ri ⊦ ri[key] = (ri \ (ks0 ∪ ks1))[key], key ∈ ks1, key ∈ ks0

Rule argument

Additional information

Used simplifier rules: unc: ⊦ ks ∪ ks0 = ks0 ∪ ks; from specification set-union[node-index]%una: ⊦ (ks ∪ ks0) ∪ ks1 = ks ∪ ks0 ∪ ks1; from specification set-union[node-index]% ⊦ ¬ key ∈ ks → (key ∈ ks ∪ ks0 ↔ key ∈ ks0); (s)% ⊦ ¬ key ∈ ks → (ri \ ks)[key] = ri[key]; (s)% ⊦ key ∈ ri \ ks ↔ key ∈ ri ∧ ¬ key ∈ ks; (ws)