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Proof step 4 from proof Diff-union-comm (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 5:
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])
Proof step 8:
⊦ 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])

Rule argument

fma: key ∈ (ri \ ks0) \ ks1 ↔ key ∈ ri \ (ks0 ∪ ks1)

Additional information

Used formulas: key ∈ (ri \ ks0) \ ks1 ↔ key ∈ ri \ (ks0 ∪ ks1)