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Proof step 62 from proof program+leftmost (previous step)

r ≠ null, (r => mknode(r0, a0, r1) * tr(r0, t0) * tr(r1, t1) * P)(H), r2 = H[r] .left ⊦ 〈H0 := H, r0 := r〉〈if r2 = null then { a1 := H0[r0] .val ; r0 := H0[r0] .right } else let r = r0, r0 = H0[r2] .left in { while r0 ≠ null do { r := r2 ; r2 := r0 ; r0 := H0[r2] .left } ; a1 := H0[r2] .val ; r0 := H0[r2] .right ; H0[r] .left := r0 }〉 ((tr(r0, (t0 = leaf ⊃ t1; branch(t0 .butleftmost, a0, t1))) * P * true)(H0) ∧ a1 = ((t0 = leaf ⊃ a0; t0 .leftmost)))

Heuristic symbolic execution applied rule assign right

And obtained the following new goals

Proof step 61:
r ≠ null, (r => mknode(r0, a0, r1) * tr(r0, t0) * tr(r1, t1) * P)(H), r2 = H[r] .left, H1 = H ∧ r3 = r ⊦ 〈if r2 = null then { a1 := H1[r3] .val ; r3 := H1[r3] .right } else let r = r3, r0 = H1[r2] .left in { while r0 ≠ null do { r := r2 ; r2 := r0 ; r0 := H1[r2] .left } ; a1 := H1[r2] .val ; r0 := H1[r2] .right ; H1[r] .left := r0 }〉 ((tr(r3, (t0 = leaf ⊃ t1; branch(t0 .butleftmost, a0, t1))) * P * true)(H1) ∧ a1 = ((t0 = leaf ⊃ a0; t0 .leftmost)))

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]