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Proof step 22 from proof diff-at (previous step)

r ∈ H0 ⊦ (H0 \ H1)[r] = H0[r], r ∈ H1

User applied rule induction

And obtained the following new goals

Proof step 21:
r ∈ H0, ∀ H, r, H0. # H < # H1 → (r ∈ H0 → (H0 \ H)[r] = H0[r] ∨ r ∈ H) ⊦ (H0 \ H1)[r] = H0[r] ∨ r ∈ H1

Rule argument

[INDHYPARG : [STANDARDINDUCTIONTYPE : ], true, false, # H1, [] → [], <]

Additional information

[[CONCRETESIMPRULES : ([CSIMPLESSPRED : <])], [EXTRATERMS : (# H1)]]