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Proof step 22 from proof all-classes-02 (previous step)

rk ∈ rks, is_fskey(rk0 .key), rk0 ∈ rks, # str ≤ # rk0 .key .fs .class .class, ∀ rk0. rk0 ≠ rk ∧ rk0 ∈ rks ∧ is_fskey(rk0 .key) → # rk0 .key .fs .class .class < # str, ∀ rks0. # rks0 < # rks → (true → (∃ str. ∀ rk. rk ∈ rks0 ∧ is_fskey(rk .key) → ¬ # str ≤ # rk .key .fs .class .class)) ⊦ # rks = 0, rks = ∅, ∃ str. ∀ rk. rk ∈ rks ∧ is_fskey(rk .key) → ¬ # str ≤ # rk .key .fs .class .class

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 21:
rk0 = r -- sk, rk ∈ rks, is_fskey(sk), rk0 ∈ rks, # str ≤ # sk .fs .class .class, ∀ rk0. rk0 ≠ rk ∧ rk0 ∈ rks ∧ is_fskey(rk0 .key) → # rk0 .key .fs .class .class < # str, ∀ rks0. # rks0 < # rks → (true → (∃ str. ∀ rk. rk ∈ rks0 ∧ is_fskey(rk .key) → ¬ # str ≤ # rk .key .fs .class .class)) ⊦ # rks = 0, rks = ∅, ∃ str. ∀ rk. rk ∈ rks ∧ is_fskey(rk .key) → ¬ # str ≤ # rk .key .fs .class .class

Rule argument

[speclemmaarg: ,r = rk .ref ∧ sk = rk .key ↔ rk = r -- sk ,[rk, r, sk] → [rk0, r, sk] ,T ,refkey , ,elim ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : ⊦ r = rk .ref ∧ sk = rk .key ↔ rk = r -- sk])], [EXTRATERMS : (rk0, r, sk)]]