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Proof step 7 from proof ri-cons-subseteq (previous step)

key-cons(key, fs0[ri[key]]), ri[key] ∈ fs0, key ∈ ri, ri[key] ∈ fs1, ∀ key. key ∈ ri → ri[key] ∈ fs0 ∧ key-cons(key, fs0[ri[key]]), ∀ adr. adr ∈ fs0 → adr ∈ fs1 ∧ fs0[adr] = fs1[adr] ⊦ key-cons(key, fs1[ri[key]])

User applied rule all left

And obtained the following new goals

Proof step 8:
key-cons(key, fs0[ri[key]]), ri[key] ∈ fs1, key ∈ ri, ri[key] ∈ fs0, key ∈ ri → ri[key] ∈ fs0 ∧ key-cons(key, fs0[ri[key]]), ∀ adr. adr ∈ fs0 → adr ∈ fs1 ∧ fs0[adr] = fs1[adr] ⊦ key-cons(key, fs1[ri[key]])

Rule argument

pos: left 5 terms: key abort: F discard: T computed: T

Additional information

Used terms: key