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Proof step 25 from proof all-classes-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 24:
rk ∈ rks, ∀ 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 = ∅, # rks = 0, ∃ str. ∀ rk. rk ∈ rks ∧ is_fskey(rk .key) → ¬ # str ≤ # rk .key .fs .class .class

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ¬ m ≤ n ↔ n < m], [CSIMPSEQ : ⊦ rk ∈ rks -- rk0 ↔ ¬ ¬ (rk ≠ rk0 ∧ rk ∈ rks)])]]