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Proof step 7 from proof referenced-adrs-in (previous step)

∀ adr, ri. adr ∈ referenced-adrs(ax0, ri) ↔ adr ∈ ax0 ∧ adr ∈ ri ⊦ adr ∈ referenced-adrs(ax0, ri) ↔ (adr = adr0 ∨ adr ∈ ax0) ∧ adr ∈ ri, adr ∈ (adr0 ∈ ri ⊃ adr0 ';[])

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 8:
adr ∈ referenced-adrs(ax0, ri), ∀ adr, ri. adr ∈ referenced-adrs(ax0, ri) ↔ adr ∈ ax0 ∧ adr ∈ ri ⊦ adr ∈ referenced-adrs(ax0, ri) ↔ (adr = adr0 ∨ adr ∈ ax0) ∧ adr ∈ ri, adr ∈ (adr0 ∈ ri ⊃ adr0 ';[])
Proof step 22:
∀ adr, ri. adr ∈ referenced-adrs(ax0, ri) ↔ adr ∈ ax0 ∧ adr ∈ ri ⊦ adr ∈ referenced-adrs(ax0, ri) ↔ (adr = adr0 ∨ adr ∈ ax0) ∧ adr ∈ ri, adr ∈ (adr0 ∈ ri ⊃ adr0 ';[]), adr ∈ referenced-adrs(ax0, ri)

Rule argument

fma: adr ∈ referenced-adrs(ax0, ri)

Additional information

Used formulas: adr ∈ referenced-adrs(ax0, ri)