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Proof step 18 from proof referenced-adrs-split-one (previous step)

adr1 ' + ax3 = ax + referenced-adrs(ax3, ri) + ax0, referenced-adrs(ax0, ri) = [], referenced-adrs(ax, ri) = [], referenced-adrs(ax3, ri) = adr ', adr ∈ ri, adr1 ∈ ax0 ∨ adr1 ∈ ax ⊦ adr0 ∈ ri, adr1 ∈ ri, adr1 ∈ referenced-adrs(ax3, ri), ∃ ax, ax0. adr0 ' + adr1 ' + ax3 = ax + referenced-adrs(ax3, ri) + ax0 ∧ referenced-adrs(ax, ri) = [] ∧ referenced-adrs(ax0, ri) = []

User applied rule exists right

And closed the goal!

Rule argument

pos: right 4 terms: adr0 + ax, ax0 abort: F discard: T computed: F

Additional information

Used simplifier rules: referenced-adrs-rec: ⊦ referenced-adrs(adr ' + ax, ri) = (adr ∈ ri ⊃ adr ';[]) + referenced-adrs(ax, ri);% ⊦ adr + ax = adr ' + ax; (s)%a: ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1; from specification list[address-list-set]% ⊦ adr ' + ax = adr ' + ax0 ↔ ax = ax0; (s)% ⊦ [] + ax = ax; (s) Used terms: adr0 + ax, ax0