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Proof step 11 from proof referenced-adrs-in (previous step)

adr0 ∈ ri, adr ∈ referenced-adrs(ax0, ri), ∀ adr, ri. adr ∈ referenced-adrs(ax0, ri) ↔ adr ∈ ax0 ∧ adr ∈ ri ⊦ adr = adr0, adr ∈ ax0 ∧ adr ∈ ri

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 12:
adr ∈ referenced-adrs(ax0, ri), adr0 ∈ ri, adr ∈ ax0, ∀ adr, ri. adr ∈ referenced-adrs(ax0, ri) ↔ adr ∈ ax0 ∧ adr ∈ ri ⊦ adr = adr0, adr ∈ ax0 ∧ adr ∈ ri
Proof step 14:
adr ∈ referenced-adrs(ax0, ri), adr0 ∈ ri, ∀ adr, ri. adr ∈ referenced-adrs(ax0, ri) ↔ adr ∈ ax0 ∧ adr ∈ ri ⊦ adr = adr0, adr ∈ ax0, adr ∈ ax0 ∧ adr ∈ ri

Rule argument

fma: adr ∈ ax0

Additional information

Used formulas: adr ∈ ax0