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Proof step 8 from proof nds-split (previous step)

gndx0.nds = ndx + ndx1, ∀ ndx0, ndx1. gndx0.nds = ndx0 + ndx1 → (∃ gndx2, gndx1. gndx0 = gndx2 + gndx1 ∧ ndx0 = gndx2.nds ∧ ndx1 = gndx1.nds) ⊦ ∃ gndx, gndx1. mkgnode(nd0, m, boolvar, boolvar0) ' + gndx0 = gndx + gndx1 ∧ nd0 ' + ndx = gndx.nds ∧ ndx1 = gndx1.nds

User applied rule all left

And obtained the following new goals

Proof step 9:
gndx0.nds = ndx + ndx1, gndx0.nds = ndx + ndx1 → (∃ gndx2, gndx1. gndx0 = gndx2 + gndx1 ∧ ndx = gndx2.nds ∧ ndx1 = gndx1.nds) ⊦ ∃ gndx, gndx1. mkgnode(nd0, m, boolvar, boolvar0) ' + gndx0 = gndx + gndx1 ∧ nd0 ' + ndx = gndx.nds ∧ ndx1 = gndx1.nds

Rule argument

pos: left 2 terms: ndx, ndx1 abort: F discard: T computed: T

Additional information

Used terms: ndx, ndx1