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Proof step 16 from proof referenced-adrs-split (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 17:
referenced-adrs(ax2, 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 ∈ ri, ∃ ax, ax3. adr ' + ax2 = ax + ax3 ∧ referenced-adrs(ax, ri) = ax0 ∧ referenced-adrs(ax3, ri) = ax1

Rule argument

pos: left 2 terms: ri, ax0, ax1 abort: F discard: T computed: T

Additional information

Used terms: ri, ax0, ax1