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Proof step 20 from proof diff-at (previous step)

r ∈ H0, ∀ H, r, H0. # H < # H1 → (r ∈ H0 → (H0 \ H)[r] = H0[r] ∨ r ∈ H) ⊦ (H0 \ H1)[r] = H0[r], r ∈ H1

User applied rule cut formula

And closed the goal!

Rule argument

[fma-arg: H1 = ∅]

Additional information

[[CONCRETESIMPRULES : ([CSIMPNAMEDSEQ : diff-emp, , , ⊦ H0 \ ∅ = H0])], [EXTRAFMAS : (H1 = ∅)]]