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Proof step 3 from proof referenced-adrs-in (previous step)

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

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 4:
adr ∈ (adr0 ∈ ri ⊃ adr0 ';[]), ∀ adr, ri. adr ∈ referenced-adrs(ax0, ri) ↔ adr ∈ ax0 ∧ adr ∈ ri ⊦ adr ∈ (adr0 ∈ ri ⊃ adr0 ';[]) ∨ adr ∈ referenced-adrs(ax0, ri) ↔ (adr = adr0 ∨ adr ∈ ax0) ∧ adr ∈ ri
Proof step 6:
∀ adr, ri. adr ∈ referenced-adrs(ax0, ri) ↔ adr ∈ ax0 ∧ adr ∈ ri ⊦ adr ∈ (adr0 ∈ ri ⊃ adr0 ';[]) ∨ adr ∈ referenced-adrs(ax0, ri) ↔ (adr = adr0 ∨ adr ∈ ax0) ∧ adr ∈ ri, adr ∈ (adr0 ∈ ri ⊃ adr0 ';[])

Rule argument

fma: adr ∈ (adr0 ∈ ri ⊃ adr0 ';[])

Additional information

Used formulas: adr ∈ (adr0 ∈ ri ⊃ adr0 ';[])