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Proof step 25 from proof replayorphans-prepost (previous step)

ro1.inodes?, ks0.inodes?, key ∈ ks0, key ∈ ro1, ks0 ⊆ ro1, ∀ key. (key ∈ ri0 ⊖ ks0 ↔ key ∈ ri1 ⊖ ro1) ∧ (ri0 ⊖ ks0)[key] = (ri1 ⊖ ro1)[key] ⊦ # ks0 = 0, # ro1 = 0, ks0 = ∅, ro1 = ∅, (key0 ∈ ((ri0 \ datakeys(key, ri0)) -- key) ⊖ (ks0 -- key) ↔ key0 ∈ ri1 ⊖ ro1) ∧ (((ri0 \ datakeys(key, ri0)) -- key) ⊖ (ks0 -- key))[key0] = (ri1 ⊖ ro1)[key0]

User applied rule all left

And obtained the following new goals

Proof step 26:
key0 ∈ ri0 ⊖ ks0 ↔ key0 ∈ ri1 ⊖ ro1, (ri0 ⊖ ks0)[key0] = (ri1 ⊖ ro1)[key0], ks0.inodes?, ro1.inodes?, key ∈ ro1, key ∈ ks0, ks0 ⊆ ro1, ∀ key. (key ∈ ri0 ⊖ ks0 ↔ key ∈ ri1 ⊖ ro1) ∧ (ri0 ⊖ ks0)[key] = (ri1 ⊖ ro1)[key] ⊦ # ro1 = 0, # ks0 = 0, ks0 = ∅, ro1 = ∅, (key0 ∈ ((ri0 \ datakeys(key, ri0)) -- key) ⊖ (ks0 -- key) ↔ key0 ∈ ri1 ⊖ ro1) ∧ (((ri0 \ datakeys(key, ri0)) -- key) ⊖ (ks0 -- key))[key0] = (ri1 ⊖ ro1)[key0]

Rule argument

pos: left 6 terms: key0 abort: F discard: F computed: T

Additional information

Used terms: key0