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Proof step 21 from proof program+leftmost (previous step)

∀ t0, r5, r, r0, r3, a, r6, t2, r1, t1, a0, P, H. # t0 < # leaf → ( r5 = r ∧ r0 ≠ null ∧ r ≠ null ∧ (r0 => mknode(r3, a, r6) * tr(r3, t0) * tr(r6, t2) * tr(r1, t1) * r => mknode(r0, a0, r1) * P)(H) → 〈while r3 ≠ null do { r5 := r0 ; r0 := r3 ; r3 := H[r0] .left }〉〈a1 := H[r0] .val ; r3 := H[r0] .right ; H[r5] .left := r3〉 ((tr(r, branch((t0 = leaf ⊃ t2; branch(t0 .butleftmost, a, t2)), a0, t1)) * P * true)(H) ∧ a1 = ((t0 = leaf ⊃ a; t0 .leftmost)))), r5 = r, r3 = null, r0 ≠ null, r ≠ null, ((((r => mknode(r0, a0, r1) * r0 => mknode(null, a, r6)) * tr(r1, t1)) * tr(r6, t2)) * P)(H) ⊦ 〈H0 := H[r5, H[r5] .left:= H[r0] .right]〉 ((tr(r, branch(t2, a0, t1)) * P * true)(H0) ∧ H[r0] .val = a)

Heuristic symbolic execution applied rule assign right

And obtained the following new goals

Proof step 20:
∀ t0, r5, r, r0, r3, a, r6, t2, r1, t1, a0, P, H. # t0 < # leaf → ( r5 = r ∧ r0 ≠ null ∧ r ≠ null ∧ (r0 => mknode(r3, a, r6) * tr(r3, t0) * tr(r6, t2) * tr(r1, t1) * r => mknode(r0, a0, r1) * P)(H) → 〈while r3 ≠ null do { r5 := r0 ; r0 := r3 ; r3 := H[r0] .left }〉〈a1 := H[r0] .val ; r3 := H[r0] .right ; H[r5] .left := r3〉 ((tr(r, branch((t0 = leaf ⊃ t2; branch(t0 .butleftmost, a, t2)), a0, t1)) * P * true)(H) ∧ a1 = ((t0 = leaf ⊃ a; t0 .leftmost)))), r5 = r, r3 = null, r0 ≠ null, r ≠ null, ((((r => mknode(r0, a0, r1) * r0 => mknode(null, a, r6)) * tr(r1, t1)) * tr(r6, t2)) * P)(H), H1 ≡ H[r5, H[r5] .left:= H[r0] .right], ¬ (tr(r, branch(t2, a0, t1)) * P * true)(H1) ∨ H[r0] .val ≠ a ⊦

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]