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Proof step 9 from proof referenced-adrs-in (previous step)

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

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 10:
adr ∈ referenced-adrs(ax0, ri), adr0 ∈ ri, ∀ adr, ri. adr ∈ referenced-adrs(ax0, ri) ↔ adr ∈ ax0 ∧ adr ∈ ri ⊦ adr ∈ (adr0 ∈ ri ⊃ adr0 ';[]), (adr = adr0 ∨ adr ∈ ax0) ∧ adr ∈ ri
Proof step 16:
adr ∈ referenced-adrs(ax0, ri), ∀ adr, ri. adr ∈ referenced-adrs(ax0, ri) ↔ adr ∈ ax0 ∧ adr ∈ ri ⊦ adr ∈ (adr0 ∈ ri ⊃ adr0 ';[]), adr0 ∈ ri, (adr = adr0 ∨ adr ∈ ax0) ∧ adr ∈ ri

Rule argument

fma: adr0 ∈ ri

Additional information

Used formulas: adr0 ∈ ri