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Proof step 10 from proof referenced-adrs-split (previous step)

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) ⊦ ∃ ax, ax3. adr ' + ax2 = ax + ax3 ∧ referenced-adrs(ax, ri) = [] ∧ referenced-adrs(ax3, ri) = adr ' + referenced-adrs(ax2, ri)

User applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: [], adr ' + ax2 abort: F discard: T computed: F

Additional information

Used simplifier rules: referenced-adrs-base: ⊦ referenced-adrs([], ri) = [];%referenced-adrs-rec: ⊦ referenced-adrs(adr ' + ax, ri) = (adr ∈ ri ⊃ adr ';[]) + referenced-adrs(ax, ri);% ⊦ [] + ax = ax; (s)%a: ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1; from specification list[address-list-set]% ⊦ adr ' + ax = adr ' + ax0 ↔ ax = ax0; (s) Used terms: [], adr ' + ax2