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Proof step 8 from proof referenced-adrs-split (previous step)

adr ' + referenced-adrs(ax2, ri) = ax0 + ax1, adr ∈ ri, ∀ ri, ax0, ax1. referenced-adrs(ax2, ri) = ax0 + ax1 → (∃ ax4, ax3. ax2 = ax4 + ax3 ∧ referenced-adrs(ax4, ri) = ax0 ∧ referenced-adrs(ax3, ri) = ax1), adr ∈ ax0 ∨ adr ∈ ax1 ⊦ ∃ ax, ax3. adr ' + ax2 = ax + ax3 ∧ referenced-adrs(ax, ri) = ax0 ∧ referenced-adrs(ax3, ri) = ax1

User applied rule constructor cut

And obtained the following new goals

Proof step 9:
adr ' + referenced-adrs(ax2, ri) = ax0 + ax1, ax0 = [], adr ∈ ri, ∀ ri, ax0, ax1. referenced-adrs(ax2, ri) = ax0 + ax1 → (∃ ax4, ax3. ax2 = ax4 + ax3 ∧ referenced-adrs(ax4, ri) = ax0 ∧ referenced-adrs(ax3, ri) = ax1), adr ∈ ax0 ∨ adr ∈ ax1 ⊦ ∃ ax, ax3. adr ' + ax2 = ax + ax3 ∧ referenced-adrs(ax, ri) = ax0 ∧ referenced-adrs(ax3, ri) = ax1
Proof step 11:
adr ' + referenced-adrs(ax2, ri) = ax0 + ax1, ax0 = adr0 + ax5, adr ∈ ri, ∀ ri, ax0, ax1. referenced-adrs(ax2, ri) = ax0 + ax1 → (∃ ax4, ax3. ax2 = ax4 + ax3 ∧ referenced-adrs(ax4, ri) = ax0 ∧ referenced-adrs(ax3, ri) = ax1), adr ∈ ax0 ∨ adr ∈ ax1 ⊦ ∃ ax, ax3. adr ' + ax2 = ax + ax3 ∧ referenced-adrs(ax, ri) = ax0 ∧ referenced-adrs(ax3, ri) = ax1

Rule argument

var: ax0

Additional information

Used simplifier rules: address-list freely generated by [], +;;