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Proof step 20 from proof all-classes-02 (previous step)

rk ∈ rks, is_fskey(sk), r -- sk ∈ 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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 19:
sk = fs ', rk ∈ rks, is_fskey(sk), r -- sk ∈ rks, # str ≤ # 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: ,is_fskey(sk) → (fs = sk .fs ↔ sk = fs ') ,[sk, fs] → [sk, fs] ,T ,storekey , ,elim-03 ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : ⊦ is_fskey(sk) → (fs = sk .fs ↔ sk = fs ')])], [EXTRATERMS : (sk, fs)]]