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Proof step 19 from proof icrash-remove-refsize0-prefix (previous step)

istore0[n] = indexblock(indx, lax, m), istore[n] = indexblock(indx0, lax0, m0), n ∈ istore0, n ∈ istore, dom istore0 ⊆ dom istore, ∀ n. n ∈ istore0 ∧ n ∈ istore → istore0[n].indlist ⊑ istore[n].indlist ∧ istore0[n].ladrlist ⊑ istore[n].ladrlist ∧ # istore0[n].indlist = # istore0[n].ladrlist ∧ # istore[n].indlist = # istore[n].ladrlist, ∀ n. (n ∈ istore0 ↔ n ∈ remove-refsize0(istore0)) ∧ istore0[n] = remove-refsize0(istore0)[n] ⊦ m = 0, indx ⊑ indx0 ∧ lax ⊑ lax0 ∧ # indx = # lax ∧ # indx0 = # lax0

User applied rule all left

And closed the goal!

Rule argument

pos: left 6 terms: n abort: F discard: T computed: T

Additional information

Used simplifier rules: ⊦ indexblock(indx, lax, m).indlist = indx; (s)% ⊦ indexblock(indx, lax, m).ladrlist = lax; (s) Used terms: n