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Proof step 9 from proof in-del-01 (previous step)

∀ r1, sk0. r0 ≠ r1 → (r1 -- sk0 ∈ st ↔ r1 -- sk0 ∈ st -- r0) ⊦ r -- sk ∈ st -- r0 ↔ r ≠ r0 ∧ r -- sk ∈ st, r0 ∈ st -- r0

User applied rule case distinction

And obtained the following new goals

Proof step 5:
r -- sk ∈ st -- r0, ∀ r1, sk0. r0 ≠ r1 → (r1 -- sk0 ∈ st ↔ r1 -- sk0 ∈ st -- r0) ⊦ r ≠ r0 ∧ r -- sk ∈ st, r0 ∈ st -- r0
Proof step 8:
r ≠ r0 ∧ r -- sk ∈ st, ∀ r1, sk0. r0 ≠ r1 → (r1 -- sk0 ∈ st ↔ r1 -- sk0 ∈ st -- r0) ⊦ r -- sk ∈ st -- r0, r0 ∈ st -- r0

Rule argument

[CASEDARG : pos: right 1, ()]

Additional information

[[CONCRETESIMPRULES : ()]]