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Proof step 3 from proof in-keys-elim (previous step)

∀ key, fs. key ∈ keys(fs, ax0) ↔ (∃ adr. adr ∈ ax0 ∧ fs[adr].key = key) ⊦ key = fs[adr].key ∨ key ∈ keys(fs, ax0) ↔ (∃ adr0. (adr0 = adr ∨ adr0 ∈ ax0) ∧ fs[adr0].key = key)

Heuristic quantifier applied rule all left

And obtained the following new goals

Proof step 4:
key ∈ keys(fs, ax0) ↔ (∃ adr. adr ∈ ax0 ∧ fs[adr].key = key), ∀ key, fs. key ∈ keys(fs, ax0) ↔ (∃ adr. adr ∈ ax0 ∧ fs[adr].key = key) ⊦ key = fs[adr].key ∨ key ∈ keys(fs, ax0) ↔ (∃ adr0. (adr0 = adr ∨ adr0 ∈ ax0) ∧ fs[adr0].key = key)

Rule argument

pos: left 1 terms: key, fs

Additional information

Used terms: key, fs