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Proof step 41 from proof in-put (previous step)

sdt0[str0] = dnode(md, sdt), str0 ∈ sdt0, (str ' ⊑ q0 ⊃ q0.tail ∈ tree0;q0 ∈ dnode(md, sdt)), ∀ md1, sdt1, md2, sdt2, q, str2, tree. dnode(md1, sdt1) = dnode(md2, sdt2) → (q ∈ dnode(md2, sdt2[str2, tree]) ↔ (str2 ' ⊑ q ⊃ q.tail ∈ tree;q ∈ dnode(md1, sdt1))) ⊦ q0 ∈ dnode(md, sdt[str, tree0])

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 42:
sdt0[str0] = dnode(md, sdt), str0 ∈ sdt0, q0.tail ∈ tree0, str ' ⊑ q0, ∀ md1, sdt1, md2, sdt2, q, str2, tree. dnode(md1, sdt1) = dnode(md2, sdt2) → (q ∈ dnode(md2, sdt2[str2, tree]) ↔ (str2 ' ⊑ q ⊃ q.tail ∈ tree;q ∈ dnode(md1, sdt1))) ⊦ q0 ∈ dnode(md, sdt[str, tree0])
Proof step 44:
sdt0[str0] = dnode(md, sdt), str0 ∈ sdt0, q0 ∈ dnode(md, sdt), ∀ md1, sdt1, md2, sdt2, q, str2, tree. dnode(md1, sdt1) = dnode(md2, sdt2) → (q ∈ dnode(md2, sdt2[str2, tree]) ↔ (str2 ' ⊑ q ⊃ q.tail ∈ tree;q ∈ dnode(md1, sdt1))) ⊦ q0 ∈ dnode(md, sdt[str, tree0]), str ' ⊑ q0

Rule argument

pos: left 3

Proof step 41 from proof in-put (previous step)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

Additional information