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Proof step 12 from proof referenced-adrs-split-one (previous step)

(adr1 ∈ ri ⊃ adr1 ';[]) + referenced-adrs(ax3, ri) = adr ', adr ∈ ri, ∀ ri0, adr2. (adr1 ∈ ri0 ⊃ adr1 ';[]) + referenced-adrs(ax3, ri0) = adr2 ' → (∃ ax, ax0. adr1 ' + ax3 = ax + ((adr1 ∈ ri0 ⊃ adr1 ';[]) + referenced-adrs(ax3, ri0)) + ax0 ∧ referenced-adrs(ax, ri0) = [] ∧ referenced-adrs(ax0, ri0) = [] ∧ adr2 ∈ ri0) ⊦ adr0 ∈ ri, ∃ ax, ax0. adr0 ' + adr1 ' + ax3 = ax + adr ' + ax0 ∧ referenced-adrs(ax, ri) = [] ∧ referenced-adrs(ax0, ri) = []

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 13:
(adr1 ∈ ri ⊃ adr1 ';[]) + referenced-adrs(ax3, ri) = adr ', adr ∈ ri, adr1 ∈ ri, ∀ ri0, adr2. (adr1 ∈ ri0 ⊃ adr1 ';[]) + referenced-adrs(ax3, ri0) = adr2 ' → (∃ ax, ax0. adr1 ' + ax3 = ax + ((adr1 ∈ ri0 ⊃ adr1 ';[]) + referenced-adrs(ax3, ri0)) + ax0 ∧ referenced-adrs(ax, ri0) = [] ∧ referenced-adrs(ax0, ri0) = [] ∧ adr2 ∈ ri0) ⊦ adr0 ∈ ri, ∃ ax, ax0. adr0 ' + adr1 ' + ax3 = ax + adr ' + ax0 ∧ referenced-adrs(ax, ri) = [] ∧ referenced-adrs(ax0, ri) = []
Proof step 15:
(adr1 ∈ ri ⊃ adr1 ';[]) + referenced-adrs(ax3, ri) = adr ', adr ∈ ri, ∀ ri0, adr2. (adr1 ∈ ri0 ⊃ adr1 ';[]) + referenced-adrs(ax3, ri0) = adr2 ' → (∃ ax, ax0. adr1 ' + ax3 = ax + ((adr1 ∈ ri0 ⊃ adr1 ';[]) + referenced-adrs(ax3, ri0)) + ax0 ∧ referenced-adrs(ax, ri0) = [] ∧ referenced-adrs(ax0, ri0) = [] ∧ adr2 ∈ ri0) ⊦ adr0 ∈ ri, adr1 ∈ ri, ∃ ax, ax0. adr0 ' + adr1 ' + ax3 = ax + adr ' + ax0 ∧ referenced-adrs(ax, ri) = [] ∧ referenced-adrs(ax0, ri) = []

Rule argument

fma: adr1 ∈ ri

Additional information

Used formulas: adr1 ∈ ri