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Proof step 4 from proof referenced-adrs-split-one (previous step)

(adr0 ∈ ri ⊃ adr0 ';[]) + referenced-adrs(ax0, ri) = adr ', ∀ ri, adr. referenced-adrs(ax0, ri) = adr ' → (∃ ax2, ax1. ax0 = ax2 + referenced-adrs(ax0, ri) + ax1 ∧ referenced-adrs(ax2, ri) = [] ∧ referenced-adrs(ax1, ri) = [] ∧ adr ∈ ri) ⊦ ∃ ax, ax1. adr0 ' + ax0 = ax + adr ' + ax1 ∧ referenced-adrs(ax, ri) = [] ∧ referenced-adrs(ax1, ri) = [] ∧ adr ∈ ri

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 5:
(adr0 ∈ ri ⊃ adr0 ';[]) + referenced-adrs(ax0, ri) = adr ', adr0 ∈ ri, ∀ ri, adr. referenced-adrs(ax0, ri) = adr ' → (∃ ax2, ax1. ax0 = ax2 + referenced-adrs(ax0, ri) + ax1 ∧ referenced-adrs(ax2, ri) = [] ∧ referenced-adrs(ax1, ri) = [] ∧ adr ∈ ri) ⊦ ∃ ax, ax1. adr0 ' + ax0 = ax + adr ' + ax1 ∧ referenced-adrs(ax, ri) = [] ∧ referenced-adrs(ax1, ri) = [] ∧ adr ∈ ri
Proof step 7:
(adr0 ∈ ri ⊃ adr0 ';[]) + referenced-adrs(ax0, ri) = adr ', ∀ ri, adr. referenced-adrs(ax0, ri) = adr ' → (∃ ax2, ax1. ax0 = ax2 + referenced-adrs(ax0, ri) + ax1 ∧ referenced-adrs(ax2, ri) = [] ∧ referenced-adrs(ax1, ri) = [] ∧ adr ∈ ri) ⊦ adr0 ∈ ri, ∃ ax, ax1. adr0 ' + ax0 = ax + adr ' + ax1 ∧ referenced-adrs(ax, ri) = [] ∧ referenced-adrs(ax1, ri) = [] ∧ adr ∈ ri

Rule argument

fma: adr0 ∈ ri

Additional information

Used formulas: adr0 ∈ ri