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Proof step 6 from proof keys-in-elim-last-oc (previous step)

key ∈ keys(fs, ax0) → ax0 = ax + adr0 ' + ax1 ∧ fs[adr0].key = key ∧ ¬ fs[adr0].key ∈ keys(fs, ax1), key = fs[adr].key ∨ key ∈ keys(fs, ax0) ⊦ ∃ ax, ax1, adr0. adr ' + ax0 = ax + adr0 ' + ax1 ∧ fs[adr0].key = key ∧ ¬ fs[adr0].key ∈ keys(fs, ax1)

User applied rule cut formula

And obtained the following new goals

Proof step 7:
key ∈ keys(fs, ax0), key ∈ keys(fs, ax0) → ax0 = ax + adr0 ' + ax1 ∧ fs[adr0].key = key ∧ ¬ fs[adr0].key ∈ keys(fs, ax1), key = fs[adr].key ∨ key ∈ keys(fs, ax0) ⊦ ∃ ax, ax1, adr0. adr ' + ax0 = ax + adr0 ' + ax1 ∧ fs[adr0].key = key ∧ ¬ fs[adr0].key ∈ keys(fs, ax1)
Proof step 9:
key ∈ keys(fs, ax0) → ax0 = ax + adr0 ' + ax1 ∧ fs[adr0].key = key ∧ ¬ fs[adr0].key ∈ keys(fs, ax1), key = fs[adr].key ∨ key ∈ keys(fs, ax0) ⊦ key ∈ keys(fs, ax0), ∃ ax, ax1, adr0. adr ' + ax0 = ax + adr0 ' + ax1 ∧ fs[adr0].key = key ∧ ¬ fs[adr0].key ∈ keys(fs, ax1)

Rule argument

fma: key ∈ keys(fs, ax0)

Additional information

Used formulas: key ∈ keys(fs, ax0)