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Proof step 5 from proof abs-log-new (previous step)

gstore[lnum] = groupblock(gndx, lax, m), ginval ∩ elems nx0 = ∅, new(ax, nodes(gstore, ginval)), lnum ∈ gstore, elems nx0 ⊆ dom gstore, ∀ ginval, ax, gstore. elems nx0 ∩ ginval = ∅ ∧ new(ax, nodes(gstore, ginval)) ∧ elems nx0 ⊆ dom gstore → ax.toSet ∩ abs-log(nx0, gstore).toSet = ∅ ⊦ #(ax.toSet ∩ ((lax.to-adr(lnum)).toSet ∪ abs-log(nx0, gstore).toSet)) = 0, ax.toSet ∩ ((lax.to-adr(lnum)).toSet ∪ abs-log(nx0, gstore).toSet) = ∅, lnum ∈ ginval

User applied rule all left

And obtained the following new goals

Proof step 6:
gstore[lnum] = groupblock(gndx, lax, m), ginval ∩ elems nx0 = ∅, new(ax, nodes(gstore, ginval)), lnum ∈ gstore, elems nx0 ⊆ dom gstore, elems nx0 ∩ ginval = ∅ ∧ new(ax, nodes(gstore, ginval)) ∧ elems nx0 ⊆ dom gstore → ax.toSet ∩ abs-log(nx0, gstore).toSet = ∅ ⊦ #(ax.toSet ∩ ((lax.to-adr(lnum)).toSet ∪ abs-log(nx0, gstore).toSet)) = 0, ax.toSet ∩ ((lax.to-adr(lnum)).toSet ∪ abs-log(nx0, gstore).toSet) = ∅, lnum ∈ ginval

Rule argument

pos: left 6 terms: ginval, ax, gstore abort: F discard: T computed: T

Additional information

Used terms: ginval, ax, gstore