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Proof step 16 from proof nodes-cons-rec (previous step)

# ax = # ndx, gstore[@(m, m0, m1)] = mkgnode(nd0, m2, boolvar, boolvar0), ax[n] = @(m3, m4, m5), gstore[@(m3, m4, m5)] = mkgnode(nd1, m6, boolvar1, boolvar2), n < # ndx, @(m, m0, m1) ∈ gstore, ∀ n0. ¬ # ndx < n0 → (@(m, m0, m1) ' + ax)[n0] ∈ gstore ∧ ¬ (@(m, m0, m1) ' + ax)[n0].lnum ∈ ginval ∧ gstore[(@(m, m0, m1) ' + ax)[n0]].nd = (nd0 ' + ndx)[n0] ⊦ ndx = [], m ∈ ginval, @(m3, m4, m5) ∈ gstore ∧ ¬ m3 ∈ ginval ∧ nd1 = ndx[n]

User applied rule all left

And closed the goal!

Rule argument

pos: left 7 terms: n + 1 abort: F discard: T computed: F

Additional information

Used simplifier rules: ⊦ m < n + 1 ↔ ¬ n < m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ (adr ' + ax)[n + 1] = ax[n]; (s)%a: ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1; from specification list[address-list-set]% ⊦ (nd ' + ndx)[n + 1] = ndx[n]; (s)%a: ⊦ (ndx + ndx0) + ndx1 = ndx + ndx0 + ndx1; from specification list[node-list]% ⊦ @(m, m0, m1).lnum = m; (s)% ⊦ mkgnode(nd, m, boolvar, boolvar0).nd = nd; (s) Used terms: n + 1