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Proof step 3 from proof ri-cons-variant1-01 (previous step)

key ∈ ri0, key ∈ ri → ri[key] ∈ fs ∧ key-cons(key, fs[ri[key]]), ∀ key. key ∈ ri0 → ri0[key] ∈ fs0 ∧ key-cons(key, fs0[ri0[key]]), ∀ key. key ∈ ri → ri[key] ∈ fs ∧ key-cons(key, fs[ri[key]]) ⊦ ri0[key] ∈ fs0 ∧ key-cons(key, fs0[ri0[key]])

User applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: key abort: F discard: T computed: T

Additional information

Used terms: key